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.

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A377457 Numbers k such that k and k+1 are both terms in A377386.

Original entry on oeis.org

1, 12563307224, 15897851550, 30412355999, 37706988600, 52576459775, 67673545631, 118533901904, 244316235000, 297265003100, 332110595000, 340800265728, 349358409503, 375624917760, 378624889440, 416375389115, 450026519903, 561162864248, 596004199840, 728643460544
Offset: 1

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Author

Amiram Eldar, Oct 29 2024

Keywords

Examples

			12563307224 is a term since both 12563307224 and 12563307225 are in A377386: 12563307224/A034968(12563307224) = 369509036, 369509036/A034968(369509036) = 9723922 and 9723922/A034968(9723922) = 373997 are integers, and 12563307225/A034968(12563307225) = 358951635, 358951635/A034968(358951635) = 7976703 and 7976703/A034968(7976703) = 257313 are integers.
		

Crossrefs

Cf. A034968.
Subsequence of A118363, A328205, A377385, A377386 and A377455.
Analogous sequences: A376795 (binary), A377272 (Zeckendorf).

Programs

  • PARI
    fdigsum(n) = {my(k = n, m = 2, r, s = 0); while([k, r] = divrem(k, m); k != 0 || r != 0, s += r; m++); s;}
    is1(k) = {my(f = fdigsum(k), f2, m); if(k % f, return(0)); m = k/f; f2 = fdigsum(m); !(m % f2) && !((m/f2) % fdigsum(m/f2));}
    lista(kmax) = {my(q1 = is1(1), q2); for(k = 2, kmax, q2 = is1(k); if(q1 && q2, print1(k-1, ", ")); q1 = q2);}

A377456 Starts of runs of 3 consecutive integers that are all terms of A377385.

Original entry on oeis.org

39998374960, 326660221888, 520935101440, 723006782783, 923072388208, 977932351240, 1134397887874, 1351753892944, 1864828904536, 2171452161023
Offset: 1

Views

Author

Amiram Eldar, Oct 29 2024

Keywords

Examples

			39998374960 is a term since 39998374960, 39998374961 and 39998374962 are all in A377385: 39998374960/A034968(39998374960) = 999959374, and 999959374/A034968(999959374) = 32256754 are integers, 39998374961/A034968(39998374961) = 975570121, and 975570121/A034968(975570121) = 33640349 are integers, and 39998374962/A034968(39998374962) = 1025599358, and 1025599358/A034968(1025599358) = 30164687 are integers.
		

Crossrefs

Cf. A034968.
Subsequence of A118363, A328205, A377385 and A377455.
Analogous sequences: A376794 (binary), A377273 (Zeckendorf).

Programs

  • PARI
    fdigsum(n) = {my(k = n, m = 2, r, s = 0); while([k, r] = divrem(k, m); k != 0 || r != 0, s += r; m++); s;}
    is1(k) = {my(f = fdigsum(k)); !(k % f) && !((k/f) % fdigsum(k/f));}
    lista(kmax) = {my(q1 = is1(1), q2 = is1(2), q3); for(k = 3, kmax, q3 = is1(k); if(q1 && q2 && q3, print1(k-2, ", ")); q1 = q2; q2 = q3);}
Showing 1-2 of 2 results.