cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-5 of 5 results.

A091216 Numbers k such that numerator of Bernoulli(2*k) is divisible by 37, the first irregular prime.

Original entry on oeis.org

16, 34, 37, 52, 70, 74, 88, 106, 111, 124, 142, 148, 160, 178, 185, 196, 214, 222, 232, 250, 259, 268, 286, 296, 304, 322, 333, 340, 358, 370, 376, 394, 407, 412, 430, 444, 448, 466, 481, 484, 502, 518, 520, 538, 555, 556, 574, 592, 610, 628
Offset: 1

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Author

N. J. A. Sloane, Feb 24 2004

Keywords

Crossrefs

Programs

  • Mathematica
    Select[ Range[ 681], Mod[ Numerator[ BernoulliB[2# ]], 37] == 0 &] (* Robert G. Wilson v, Feb 24 2004 *)
  • PARI
    for(j=1,260, if (! (numerator(bernfrac(2*j))%37), print1(j, ", ")))

Extensions

More terms from Robert G. Wilson v, Feb 24 2004

A251782 Least even integer k such that numerator(B_k) == 0 (mod 37^n).

Original entry on oeis.org

32, 284, 37580, 1072544, 55777784, 325656968, 42764158652, 2444284077476, 46872402575720, 4093248733492712, 167845040875289732, 4841789050865438960, 235423026877046134208, 7818983737604766777920, 95503904455394036720840, 6908622244227620311285724, 114945213060615779807957456
Offset: 1

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Author

Keywords

Comments

37 is the first irregular prime. The corresponding entry for the second irregular prime 59 is A299466, and for the third irregular prime 67 is A299467.
The p-adic digits of the unique simple zero of the p-adic zeta function zeta_{(p,l)} with (p,l)=(37,32) were used to compute the sequence (see the Mathematica program below). This corresponds with Table A.2 in Kellner (2007). The sequence is increasing, but some consecutive entries are identical, e.g., entries 18 / 19 and 80 / 81. This is caused only by those p-adic digits that are zero.

Examples

			a(3) = 37580 because the numerator of B_37580 is divisible by 37^3 and there is no even integer less than 37580 for which this is the case.
		

Crossrefs

Programs

  • Mathematica
    p = 37; l = 32; LD = {7, 28, 21, 30, 4, 17, 26, 13, 32, 35, 27, 36, 32, 10, 21, 9, 11, 0, 1, 13, 6, 8, 10, 11, 10, 11, 32, 13, 30, 10, 6, 8, 2, 12, 1, 8, 2, 5, 3, 10, 19, 8, 4, 7, 19, 27, 33, 29, 29, 11, 2, 23, 8, 34, 5, 8, 35, 35, 13, 31, 29, 6, 7, 22, 13, 29, 7, 15, 22, 20, 19, 29, 2, 14, 2, 2, 31, 11, 4, 0, 27, 8, 10, 23, 17, 35, 15, 32, 22, 14, 7, 18, 8, 3, 27, 35, 33, 31, 6}; CalcIndex[L_, p_, l_, n_] := l + (p - 1) Sum[L[[i + 1]] p^i , {i, 0, n - 2}]; Table[CalcIndex[LD, p, l, n], {n, 1, Length[LD] + 1}] // TableForm

Formula

Numerator(B_{a(n)}) == 0 (mod 37^n).

Extensions

Edited for consistency with A299466 and A299467 by Bernd C. Kellner and Jonathan Sondow, Feb 20 2018

A299466 Least even integer k such that numerator(B_k) == 0 (mod 59^n).

Original entry on oeis.org

44, 914, 86464, 8162384, 436993736, 13087518620, 469209221382, 42059215391408, 4083629226737464, 498021221327673308, 5020105038665551466, 1516903461301962815624, 24254443348634296180510, 2604090699795956735657960, 252229046873638875979496022
Offset: 1

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Author

Keywords

Comments

59 is the second irregular prime. The corresponding entry for the first irregular prime 37 is A251782, and for the third irregular prime 67 is A299467.
The p-adic digits of the unique simple zero of the p-adic zeta function zeta_{(p,l)} with (p,l)=(59,44) were used to compute the sequence (see the Mathematica program below). This corresponds with Table A.2 in Kellner (2007). The sequence is increasing, but some consecutive entries are identical, e.g., entries 30 / 31 and 94 / 95. This is caused only by those p-adic digits that are zero.

Examples

			a(3) = 86464 because the numerator of B_86464 is divisible by 59^3 and there is no even integer less than 86464 for which this is the case.
		

Crossrefs

Programs

  • Mathematica
    p = 59; l = 44; LD = {15, 25, 40, 36, 18, 11, 17, 28, 58, 9, 51, 13, 25, 41, 44,17, 43, 35, 21, 10, 21, 38, 9, 12, 40, 43, 45, 30, 41, 0, 3, 25, 34, 49, 45,9, 19, 48, 57, 11, 13, 29, 28, 44, 41, 37, 33, 29, 43, 8, 57, 12, 48, 15,15, 53, 57, 16, 51, 16, 54, 30, 9, 26, 8, 49, 22, 58, 11, 42, 28, 36, 33,45, 24, 32, 18, 12, 29, 45, 40, 27, 19, 40, 41, 11, 42, 49, 35, 41, 57, 54,33, 0, 34, 34, 49, 6, 31}; CalcIndex[L_, p_, l_, n_] := l + (p - 1) Sum[L[[i + 1]] p^i , {i, 0, n -2}]; Table[CalcIndex[LD, p, l, n], {n, 1, Length[LD] + 1}] // TableForm

Formula

Numerator(B_{a(n)}) == 0 (mod 59^n).

A299467 Least even integer k such that numerator(B_k) == 0 (mod 67^n).

Original entry on oeis.org

58, 3292, 153640, 12597148, 846312184, 52715297638, 320040068824, 370475739904372, 23170872799129498, 532379740455157312, 111861518490094080436, 1314934469494256636776, 291496130251698265225984, 7852328398132458266800348, 1925603427201316655808983674
Offset: 1

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Author

Keywords

Comments

67 is the third irregular prime. The corresponding entry for the first irregular prime 37 is A251782, and for the second irregular prime 59 is A299466.
The p-adic digits of the unique simple zero of the p-adic zeta function zeta_{(p,l)} with (p,l)=(67,58) were used to compute the sequence (see the Mathematica program below). This corresponds with Table A.2 in Kellner (2007). The sequence is increasing, but some consecutive entries are identical, e.g., entries 22 / 23 and 84 / 85. This is caused only by those p-adic digits that are zero.

Examples

			a(3) = 153640 because the numerator of B_153640 is divisible by 67^3 and there is no even integer less than 153640 for which this is the case.
		

Crossrefs

Programs

  • Mathematica
    p = 67; l = 58; LD = {49, 34, 42, 42, 39, 3, 62, 57, 19, 62, 10, 36, 14, 53, 57, 16, 60, 22, 41, 21, 25, 0, 56, 21, 24, 52, 33, 28, 51, 34, 60, 8, 47, 39, 42, 33, 14, 66, 50, 48, 45, 28, 61, 50, 27, 8, 30, 59, 32, 15, 3, 1, 54, 12, 30, 20, 14, 12, 10, 49, 33, 49, 54, 13, 26, 42, 8, 58, 12, 63, 19, 16, 48, 15, 2, 13, 1, 23, 2, 44, 64, 25, 40, 0, 16, 58, 44, 31, 62, 47, 61, 46, 9, 2, 50, 1, 62, 34, 31}; CalcIndex[L_, p_, l_, n_] := l + (p - 1) Sum[L[[i + 1]] p^i , {i, 0, n - 2}]; Table[CalcIndex[LD, p, l, n], {n, 1, Length[LD] + 1}] // TableForm

Formula

Numerator(B_{a(n)}) == 0 (mod 67^n).

A090789 Even numbers n such that 37^2 (the square of the first irregular prime) divides the numerator of Bernoulli(n).

Original entry on oeis.org

284, 1184, 1616, 2516, 2738, 2948, 3848, 4280, 5180, 5476, 5612, 6512, 6944, 7844, 8214, 8276, 9176, 9608, 10508, 10940, 10952, 11840, 12272, 13172, 13604, 13690, 14504, 14936, 15836, 16268, 16428, 17168, 17600, 18500, 18932, 19166
Offset: 1

Views

Author

T. D. Noe, Feb 26 2004

Keywords

Comments

Let N(n) be the numerator of the Bernoulli number B(n). This sequence is the union of three arithmetic progressions. The first, n=284+36*37*a, follows from work by Kellner on higher-order irregular pairs. In this case, the second-order pair is (37,284) because n=284 is the smallest even n such that 37^2 | N(n). The second progression, n=37(32+36*b), follows from the first-order pair (37,32). By the Kummer congruence, 37 | N(n) for n=32+36b. By a theorem of Adams, every 37th of these numbers has another factor of 37. The third progression, n=2*37^2c, yields factors of 37^2 by Adams' theorem.

Crossrefs

Twice A092230.

Programs

  • Maple
    N:= 20000: # to get all terms <= N
    sort(convert({seq(284+36*37*k, k=0..floor((N-284)/36/37)),
    seq(1184+36*37*k, k=0..floor((N-1184)/36/37)),
    seq(2*37^2*k, k=1..floor(N/2/37^2))},list)); # Robert Israel, Aug 20 2015
  • Mathematica
    nn=10; Union[284+36*37*Range[0, 2nn], 37(32+36*Range[0, 2nn]), 2*37^2*Range[nn]]

Formula

These numbers are the union of three arithmetic progressions: 284 + 36*37*k, 32*37 + 36*37*k and 2*37^2*k.

Extensions

Definition corrected by Robert Israel, Aug 20 2015
Showing 1-5 of 5 results.