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.

User: Logan J. Kleinwaks

Logan J. Kleinwaks's wiki page.

Logan J. Kleinwaks has authored 6 sequences.

A384729 A B_2-sequence with reciprocal sum > 2.1615.

Original entry on oeis.org

1, 2, 4, 8, 13, 21, 31, 45, 66, 81, 97, 123, 148, 182, 204, 252, 291, 324, 352, 415, 486, 540, 651, 706, 792, 838, 928, 1046, 1134, 1228, 1358, 1407, 1512, 1624, 1869, 1938, 2087, 2170, 2367, 2480, 2608, 2765, 3033, 3080, 3232, 3567, 3605, 3797, 3950, 4267, 4505, 4677, 5064, 5290, 5480, 5655, 6059, 6507, 6892, 6967
Offset: 1

Author

Logan J. Kleinwaks, Jun 08 2025

Keywords

Comments

This is the B_2 sequence with largest reciprocal sum that is known to the author, as of the date of submission. The reciprocal sum of the first 1010 terms, which are given in the attached b-file, is 2.16150003... This already provably exceeds the reciprocal sum of the infinite sequence by R. Lewis (A046185), defined by greedy extension of a sequence of 68 terms (e.g., compute R. Lewis' sequence to 1600 terms, then bound the remainder using the result of B. Lindström on the maximum cardinality of B2-sequences with elements in [1, N]). By extending this sequence greedily after the first 1010 terms, a larger reciprocal sum can obviously be achieved.
This sequence first differs from A046185 at the 25th term.
This sequence was found by the author by using a combination of beam search and batch greedy algorithm, as part of an experiment to evaluate LLM code generation and mathematical reasoning (all code was written by the LLM, though with significant prompting).

Crossrefs

Cf. A046185 (R. Lewis), A005282 (Mian-Chowla).

A293542 a(1)=1; for n>1, a(n) = least integer greater than a(n-1) such that the sums of the divisors of the pairwise sums of a(1),...,a(n) are all distinct.

Original entry on oeis.org

1, 2, 4, 8, 19, 28, 56, 72, 101, 144, 202, 240, 261, 448, 511, 602, 772, 806, 1152, 1296, 1541, 1602, 2016, 2256, 2900, 3322, 3362, 3978, 4376, 5887, 6416, 7702, 8228, 8578, 11341, 11382, 13376, 13692, 16083, 16380, 16544, 17382, 22726, 24944, 26302, 27508, 30580, 33184, 34020, 37474
Offset: 1

Author

Logan J. Kleinwaks, Oct 11 2017

Keywords

Examples

			Let s(n) be the sum of the divisors of n. a(3)!=3 because s(1+3)=s(2+2)=7. a(3)=4 because s(1+1)=3, s(1+2)=4, s(1+4)=6, s(2+2)=7, s(2+4)=12, and s(4,4)=15 are all distinct.
		

Crossrefs

A293541 a(1)=1; for n>1, a(n) = least integer greater than a(n-1) such that the numbers of divisors of the pairwise sums of a(1),...,a(n) are all distinct.

Original entry on oeis.org

1, 3, 15, 909, 306915, 36181647, 25342642989
Offset: 1

Author

Logan J. Kleinwaks, Oct 11 2017

Keywords

Examples

			Let d(n) be the number of divisors of n.  Then a(2)!=2 because d(1+1)=d(1+2)=2. a(2)=3 because d(1+1)=2, d(1+3)=3, and d(3+3)=4 are all distinct.
		

Crossrefs

A293457 Primes that divide the numerator of the sum of the reciprocals of all smaller primes.

Original entry on oeis.org

2, 5, 19, 47, 79, 109, 3667387
Offset: 1

Author

Logan J. Kleinwaks, Oct 09 2017

Keywords

Comments

Exhaustive search finds no more terms among the first 10^7 primes.
Primes p that divide A024451(A000720(p)-1). - Antti Karttunen, Feb 08 2024

Examples

			Since 1/2 + 1/3 + 1/5 + 1/7 + 1/11 + 1/13 + 1/17 = 716167/510510 and 19 divides 716167, 19 is in the sequence.
Since there are no primes less than 2, the sum of their reciprocals is 0/1, and as 2 divides 0, it is therefore included as the first term of this sequence. - _Antti Karttunen_, Feb 08 2024
		

Programs

  • PARI
    lista(nn) = my(s = 0); forprime(p=2, nn, if (!(numerator(s) % p), print1(p, ", ")); s += 1/p; ); \\ Michel Marcus, Oct 09 2017, edited for the new, more inclusive definition by Antti Karttunen, Feb 08 2024

Formula

a(n) = A000040(1+A369972(n)). - Antti Karttunen, Feb 08 2024

Extensions

Relaxed the definition to include 2 as the first term - Antti Karttunen, Feb 08 2024

A271058 Least m such that n equals a sum of binomial coefficients C(m,k1)+C(m,k2)+C(m,k3)+... with 0<=k1

Original entry on oeis.org

1, 1, 2, 2, 3, 3, 3, 3, 4, 4, 4, 4, 6, 4, 4, 4, 5, 8, 9, 5, 5, 5, 6, 11, 5, 5, 5, 6, 6, 5, 5, 5, 6, 6, 6, 6, 6, 6, 19, 19, 6, 6, 6, 6, 8, 8, 6, 6, 6, 6, 6, 6, 26, 9, 9, 6, 6, 6, 29, 29, 30, 6, 6, 6, 7, 8, 10, 11, 34, 7, 7, 7, 8, 8, 37, 37, 7, 7, 7, 8, 9, 9, 9, 7, 7, 7, 8, 8
Offset: 1

Author

Logan J. Kleinwaks, Mar 29 2016

Keywords

Examples

			a(3) = 2 since 3 = C(2,0) + C(2,1) but 3 != C(1,0) or C(1,1) or C(1,0) + C(1,1).
		

Programs

  • Mathematica
    nn = 22; s = Table[Union@ Map[Total, Binomial[m, #]] &@ Rest@ Subsets@ Range[0, m], {m, nn}]; Table[SelectFirst[Range[If[n == 1, 1, Ceiling@ Log2@ n], nn], MemberQ[s[[#]], n] &], {n, 52}] /. m_ /; MissingQ@ m -> 0 (* Michael De Vlieger, Mar 30 2016, Version 10.2, values of m greater than nn show as "0". *)

A270875 Number of subsets of {1,...,n} with sum of elements equal to least common multiple of elements and at least two elements.

Original entry on oeis.org

0, 0, 1, 1, 1, 2, 2, 3, 5, 6, 6, 9, 9, 10, 13, 15, 15, 19, 19, 33, 35, 35, 35, 47, 47, 47, 50, 57, 57, 101, 101, 106, 108, 108, 108, 127, 127, 127, 128, 249, 249, 268, 268, 272, 358, 358, 358, 406, 406, 408, 409, 411, 411, 424, 424, 501, 502, 502, 502, 1190, 1190
Offset: 1

Author

Logan J. Kleinwaks, Mar 24 2016

Keywords

Comments

It appears that the sequence of n for which a(n)>a(n-1) has a large overlap with A175904.

Examples

			For n=3, the subsets of {1,2,3} with at least two elements have (sum,LCM) as follows: {1,2}->(3,2), {1,3}->(4,3), {2,3}->(5,6), {1,2,3}->(6,6). Only the last satisfies sum=LCM, so a(3)=1.
		

Crossrefs

Cf. A270970 (similar sequence counting trivial solutions).

Programs

  • Mathematica
    Table[Length[Transpose@ {Total /@ #, LCM @@@ #} /. {a_, b_} /; a != b -> Nothing &@ Rest[Subsets[Range@ n] /. {} -> Nothing]], {n, 2, 22}] (* _Michael De Vlieger, Mar 24 2016 *)
  • PARI
    a(n) = {nb = 0; S = vector(n, k, k); for (i = 0, 2^n - 1, ss = vecextract(S, i); if (vecsum(ss) == lcm(ss), nb++);); nb - n;} \\ Michel Marcus, Mar 26 2016

Formula

a(n) = A270970(n) - n. - Michel Marcus, Mar 27 2016

Extensions

More terms added using A270970 by Jinyuan Wang, May 02 2020