Problems & Puzzles: Puzzles

 Puzzle 811. Primes and Sum of consecutive triangulars-II Jean Brette sent this nice follow-up to puzzle 810: We have seen that the sums of 3 consecutive triangular numbers are equal to 1 (mod 3) and the sums of 6 triangular are equal to 2 (mod 3)   Hence, it can exist  consecutive integers N3 and N6 such that N6 = N3 + 1   examples :199 = T10 + T11 + T12 and 200 = T5 + T6+ .....+ T10 514 = T17 + T18 +T19 and 515 = T10 + T11+ .. + T15   Q1. Find more pairs of these.   It exists also  pair  (N6, N3) where  N3 = N6 + 2   example  :83 = T2 + T3 + ...+ T7 and 85 = T6 + T7 + T8 Unfortunately, 83 is prime but 85 is not.   Question 2 : Does it exist twin primes (P6, P3 ) analog to (83, 85) ?

Contributions came from Jan van Delden, Emanuel Vantieghem and Jeff Heleen.

***

Jan wrote:

Q1:

A few terms of N3:

199,514,6634,17335,225235,588754,7651234,20000179,259916599,679417210,8829513010, 23080184839,299943525619,784046867194

Q2:

If N3=N6+2 it is straightforward to show that N3=0 mod 5.

***

Emanuel wrote:

I use the following notations :
T(n)=n(n+1)/2 = the nth triangular number
U(n) = T(n)+T(n+1)+T(n+2) = (3n^2+8n+9)/2  (derived in puzzle 810)
V(m) = T(m)+T(m+1)+T(m+2)+T(m+3)+T(m+4)+T(m+5) = 3m^2+18m+35 (id).

Q1.
We should solve the equation  U(n) +1= V(m)  or :
(3n^2+8n+9)/2 + 1 = 3m^2+18m+35
This is equivalent to
n^2 + 3n - (2m^2 + 12m +20) = 0
We can consider this as a quadratic equation in the unknown  n.
This will have integer solutions iff its discriminant  (which is  9 + 4(2m^2 + 12m +20)) is the square of an integer  x.
Thus, we should solve the equation :
x^2 = 8m^2 + 48m + 89     (*)
and then put  n = (x-3)/2.  This leads us very quickly to the solutions :
m = 5, 10, 44, 73, 271, 440, 1594, 2579, 9305, 15046, 54248, 87709, 316195, 511220, ...
n = 10, 17, 65, 106, 386, 625, 2257, 3650, 13162, 21281, 76721, 124042, 447170, 722977, ...
The corresponding pairs (U(n),V(m)) of consecutive integers are :
(199, 200), (514, 515), (6634, 6635), (17335, 17336), (225235, 225236), (588754, 588755), (7651234, 7651235), (20000179, 20000180), (259916599, 259916600), (679417210, 679417211), (8829513010, 8829513011), (23080184839, 23080184840), (299943525619, 299943525620), (784046867194, 784046867195), ...
Though there are many solutions, this does not give answer whether or not there are infinitley many ones.
But, we can rewrite  (*)  in the form   x^2 - (8m^2 + 48 m +72)   = 17 or :
x^2 - 2y^2 = 17   (**)
with  y = 2m + 6.
To solve (**) we can use the well known theory of numbers of the form  a+b*Sqrt(2) (a, b integers).
There are two 'small' solutions  (x,y) = (5,2)  and  (x,y) = (7,4).
From these solutions we can obtain an infinitude of others by the following rule :
if (x,y) is a solution, then so is  (3x+4y,2x+3y)
This let us find all the solutions.

Q2
Here, we should solve the equation  V(m)+2 = U(n)  or :
n^2 + 3n - (2m^2 + 12m +22) = 0
Here also, this is solvable in integers provided there is an integer  x  such that :
x^2 =  8m^2 + 48m + 97     (***)
and  n = (x - 3)/2.
This leads us quckly to the solutions :
m = 2, 27, 172, 1017, 5942, 34647, 201952, ...
n = 6, 41, 246, 1441, 8406, 49001, 285606, ...
The corresponding pairs  (V(m), U(n))  are :
(83, 85), (2708, 2710), (91883, 91885), (3121208, 3121210), (106029083, 106029085), (3601867508, 3601867510), (122357466083, 122357466085), ...
If we work as in Q1, we should first solve
x^2 - 2 y^2 = 25,
which has the solution (x,y) = (5, 0)  and all other solutions come from the rule that  (3x+4y,2x+3y)  is also a solution.
So, in every solution  x  and  y  are seemingly divisible by  5.  Since  n = (x - 3)/2  this implies that  n  allways will be congruent to  1 mod 5.
Thus, U(n) = (3n^2 +8n + 9)/2 = 0 mod 5.
So, there is no twinprime pair solution !

N.B. The rule to be applied on  (m,n)  is (in both cases Q1 and Q2) :
if (m,n) is a solution then so is (3m+2n+9,4m+3n+15).
(easy to check).

***

Jeff Heleen wrote:

For puzzle 811:

I looked up to T=500,000,000 and found 32 instances. Primes are marked (P).

Note the pattern of the ending digits for both Q1 and Q2. -Jeff Heleen

Q1
199 (P) = T10 + T11 + T12
200 = T5 + ... + T10

514 = T17 + T18 + T19
515 = T10 + ... + T15

6634 = T65 + T66 + T67
6635 = T44 + ... + T49

17335 = T106 + T107 + T108
17336 = T73 + ... + T78

225235 = T386 + T387 + T388
225236 = T271 + ... + T276

588754 = T625 + T626 + T627
588755 = T440 + ... + T445

7651234 = T2257 + T2258 + T2259
7651235 = T1594 + ... + T1599

20000179 = T3650 + T3651 + T3652
20000180 = T2579 + ... + T2584

259916599 = T13162 + T13163 + T13164
259916600 = T9305 + ... + T9310

679417210 = T21281 + T21282 + T21283
679417211 = T15046 + ... + T15051

8829513010 = T76721 + T76722 + T76723
8829513011 = T54248 + ... + T54253

23080184839 (P) = T124042 + T124043 + T124044
23080184840 = T87709 + ... + T87714

299943525619 = T447170 + T447171 + T447172
299943525620 = T316195 + ... + T316200

784046867194 = T722977 + T722978 + T722979
784046867195 = T511220 + ... + T511225

10189250357914 = T2606305 + T2606306 + T2606307
10189250357915 = T1842934 + ... + T1842939

26634513299635 = T4213826 + T4213827 + T4213828
26634513299636 = T2979623 + ... + T2979628

346134568643335 = T15190666 + T15190667 + T15190668
346134568643336 = T10741421 + ... + T10741426

904789405320274 = T24559985 + T24559986 + T24559987
904789405320275 = T17366530 + ... + T17366535

11758386083515354 = T88537697 + T88537698 + T88537699
11758386083515355 = T62605604 + ... + T62605609

30736205267589559 = T143146090 + T143146091 + T143146092
30736205267589560 = T101219569 + ... + T101219574

399438992270878579 = T516035522 + T516035523 + T516035524
399438992270878580 = T364892215 + ... + T364892220

Q2
85 = T6 + T7 + T8
83 (P) = T2 + ... + T7

2710 = T41 + T42 + T43
2708 = T27 + ... + T32

91885 = T246 + T247 + T248
91883 = T172 + ... + T177

3121210 = T1441 + T1442 + T1443
3121208 = T1017 + ... + T1022

106029085 = T8406 + T8407 + T8408
106029083 = T5942 + ... + T5947

3601867510 = T49001 + T49002 + T49003
3601867508 = T34647 + .. + T34652

122357466085 = T285606 + T285607 + T285608
122357466083 = T201952 + ... + T201957

4156551979210 = T1664641 + T1664642 + T1664643
4156551979208 = T1177077 + ... + T1177082

141200409826885 = T9702246 + T9702247 + T9702248
141200409826883 = T6860522 + ... + T6860527

4796657382134710 = T56548841 + T56548842 + T56548843
4796657382134708 = T39986067 + ... + T39986072

162945150582753085 = T329590806 + T329590807 + T329590808
162945150582753083 = T233055892 + ... + T233055897

***

On Dec. 6, 2015, T. D. Noe wrote:

Using the numbers that Jeff Heleen computed, it is possible to write a
recursive formula for the numbers 199, 514, 6634, 17335, 225235, 588754,
7651234, 20000179, 259916599, 679417210, 8829513010,...:

x(n+1) = x(n) + 34*x(n-1) - 34*x(n-2) - x(n-3) + x(n-4) with initial terms
{199, 514, 6634, 17335, 225235}.

...

If one plots the numbers t = {199, 514, 6634, 17335, 225235, 588754,
7651234, 20000179, 259916599, 679417210, 8829513010,...} on a logarithmic
scale, then it is apparent that the numbers may satisfy a linear recurrence
equation of some order.  I wrote a Mathematica function to search for such
equations:

FindCoef[seq_List, i_, n_] := Module[{m, a, s, noSoln = {}},
a = Take[seq, {i, i + 2 n - 1}];
m = Reverse /@ Partition[Most[a], n, 1];
If[Det[m] == 0, noSoln, s = LinearSolve[m, Take[a, -n]];
If[And @@ IntegerQ /@ s, s, noSoln]]]

which analyzes a sequence seq starting a position i for a linear relation
of degree n.  A 5th-order relation works.  Mathematica reports

In[40]:= FindCoef[t, 1, 5]

Out[40]= {1, 34, -34, -1, 1}

This answer means that the numbers satisfy the relation

x(n+1) = x(n) + 34*x(n-1) - 34*x(n-2) - x(n-3) + x(n-4)

with initial terms {199, 514, 6634, 17335, 225235}.

***

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