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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A354330 Distance from the sum of the first n positive triangular numbers to the nearest triangular number.

Original entry on oeis.org

0, 0, 1, 0, 1, 1, 1, 6, 0, 6, 10, 10, 13, 10, 1, 14, 4, 21, 12, 4, 0, 1, 8, 22, 28, 1, 36, 1, 35, 30, 10, 4, 11, 10, 0, 20, 51, 41, 10, 71, 4, 62, 41, 6, 45, 75, 91, 88, 97, 85, 55, 10, 51, 100, 10, 99, 20, 124, 29, 56, 130, 90, 48, 20, 7, 10, 30, 68, 125, 136
Offset: 0

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Author

Paolo Xausa, Jun 04 2022

Keywords

Comments

a(n) = 0 for n in {0, 1, 3, 8, 20, 34} = A224421.

Examples

			a(4) = 1 because the sum of the first 4 positive triangular numbers is 1 + 3 + 6 + 10 = 20, the nearest triangular number is 21 and 21 - 20 = 1.
		

Crossrefs

Programs

  • Mathematica
    nterms=100;Table[ts=n(n+1)(n+2)/3;t=Floor[Sqrt[ts]];Abs[t^2+t-ts]/2,{n,0,nterms-1}]
  • PARI
    a(n)=my(ts=n*(n+1)*(n+2)/3,t=sqrtint(ts));abs(t^2+t-ts)/2;
    vector(100,n,a(n-1)) \\ Paolo Xausa, Jul 06 2022
    
  • Python
    from math import isqrt
    def A354330(n): return abs((m:=isqrt(k:=n*(n*(n + 3) + 2)//3))*(m+1)-k)>>1 # Chai Wah Wu, Jul 15 2022

Formula

a(n) = A053616(A000292(n)).
a(n) = abs(A000292(n) - A354329(n)).

A233327 Distance from 2^n to the nearest triangular number.

Original entry on oeis.org

0, 1, 1, 2, 1, 4, 2, 8, 3, 16, 11, 32, 1, 64, 87, 128, 167, 256, 306, 512, 500, 1024, 552, 2048, 688, 4096, 3041, 8192, 579, 16384, 20854, 32768, 37075, 65536, 55618, 131072, 37108, 262144, 222296, 524288, 147729, 1048576, 891994, 2097152, 602155, 4194304, 3523022
Offset: 0

Views

Author

Alex Ratushnyak, Feb 23 2014

Keywords

Examples

			Triangular number nearest to 2^8=256 is 253, so a(8)=256-253=3.
		

Crossrefs

Bisections: a(2n) = abs(A238455(n)), a(2n+1) = A000079(n).

Programs

  • Mathematica
    a[n_] := Module[{k, k0}, k0 = k /. FindRoot[2^n == k*(k+1)/2, {k, 2^(n/2)}, WorkingPrecision -> 20] // Round; Abs[2^n-k0*(k0+1)/2]]; Table[a[n], {n, 0, 46}] (* Jean-François Alcover, Feb 27 2014 *)

Formula

a(2*k+1) = 2^k.
Specifically, both the nearest triangular number below: A006516(n) = A000217((2^n)-1) = 2^(2n-1) - 2^(n-1) and the nearest triangular number above: A007582(n) = A000217(2^n) = 2^(2n-1) + 2^(n-1) are at the same distance from 2^(2n-1). - Antti Karttunen, Feb 26 2014
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