Problems & Puzzles: Puzzles

Puzzle 1043. Another puzzle about Keith numbers

Paolo Lava sent the following nice puzzle.

For the definition of Keith numbers have a look to A007629.
 
For n-th power analog of Keith numbers see A274769, A274770 and from A281915 to A281921.
 
Now, I search for the minimum prime p such that by applying the Keith Number Process (KNP) to p^n, with n>0, we reach p. Let us say, KNP(p^n) = p

"I take the definition from A007629:
Numbers n>9 with following property: form a sequence b(i) whose initial terms are the t digits of n, later terms given by rule that b(i) = sum of t previous terms; then n itself appears in the sequence.

E.g.
197 is a Keith number since sequence starts 1, 9, 7. Then 1+9+7=17. So 1,9,7,17. Again 9+7+17=33. So 1,9,7,17,33. Again 7+17+33=57, 17+33+57=107, 33+57+107=197 so 1,9,7,17,33,57,107,197 ... that is the number we started from.

I identify this process with the notation KNP (Keith Number Process)."


Here below the table I got for 1<n<100:

n

p

1

2

2

37

3

17

4

7

5

109

6

 

7

31

8

80051

9

71

10

97

11

107

12

13093

13

103

14

127

15

107

16

163

17

991

18

181

19

157

20

181

21

199

22

193

23

271

24

31663

25

211

26

307

27

307

28

 

29

673

30

8297

31

331

32

811

33

359

34

463

35

 

36

473741

37

421

38

2243

39

449

40

1031

41

11503

42

487

43

461

44

523

45

503

46

1171

47

1279

48

631

49

661

50

32323

51

321221

52

739

53

683

54

739

55

677

56

1571

57

719

58

1709

59

8237

60

185303

61

 

62

1759

63

827

64

829

65

422749

66

 

67

859

68

44687

69

9769

70

2099

71

991

72

108649

73

853

74

1009

75

2281

76

1093

77

1061

78

1117

79

1031

80

23687

81

 

82

1231

83

1151

84

 

85

1051

86

2591

87

1187

88

2791

89

1151

90

 

91

 

92

2803

93

3061

94

1303

95

 

96

14107

97

1237

98

3203

99

3061

100

1489


Q1. Is there any prime for n = 6, 28, 35, 61, 66, 81, 84, 90, 91, 95 ?
I guess there is always at least a p for any n but searching for it is a time consuming task for my PC.
 

Variants
KNP(p(n)) = p(n+1)
11, 13
29, 31
31, 37
97, 101
797, 809
22073, 22079
381287, 381289
E.g.  797 -> 809
7 + 9 + 7 = 23;
9 + 7 + 23 = 39;
7 + 23 + 39 = 69;
23 + 39 + 69 = 131;
39 + 69 + 131 = 239;
69 + 131 + 239 = 439;
131 + 239 + 439 = 809.
KNP(p(n)) = p(n-1)
11, 13
131, 137
6899, 6907
1963267, 1963277
KNP(p(n)+p(n+1)) = p(n+2)
2,3,5
12487, 12491, 12497
KNP(p(n)+p(n+1)) = p(n-1)
3,5,7
53,59,61
103,107,109
733, 739, 743
KNP(p(n)*p(n+1)) = p(n+2)
73,79,83
97,101,103
KNP(p(n)*p(n+1)) = p(n-1)
13,17,19
19,23,29
3433,3449,3457
KNP(p(n) U p(n+1)) = p(n+2)
2,3,5
19,23,29
163,167,173
311,313,317
30841,30851,30853
31237,31247,31249
3438973,3438091,3438103
KNP(p(n) U p(n+1)) = p(n-1)
107,109,113
 

Q2. Other values for these variants?


During the week from June26 to July 2, contributions came from Emmanuel Vantieghem, Simon Cavegn, Giorgios  Kalogeropoulos.

***

Emmanuel wrote:

I have only two answers to question 1 :
n = 6  -> p = 36013476739 (found in  A281917)
n = 28 -> p = 5318989651.

***

Simon wrote:

Q1
n:81 p:200908021
n:84 p:25471
Searched up to p=4000000007

***

Giorgios wrote:

Q1.
n=6    -> p=36013476739 (this one was already known, see A281917)
n=28  -> p=5318989651
n=61  -> p=11659149703
n=81  -> p=200908021
n=84  -> p=25471

***

 

 

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