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 1-3 of 3 results.

A125088 a(1)=1. a(n) = sum of the earlier terms equal to any exponent in the prime-factorization of n.

Original entry on oeis.org

1, 1, 2, 2, 2, 2, 2, 0, 10, 2, 2, 16, 2, 2, 2, 0, 2, 24, 2, 26, 2, 2, 2, 2, 32, 2, 0, 36, 2, 2, 2, 0, 2, 2, 2, 46, 2, 2, 2, 2, 2, 2, 2, 62, 62, 2, 2, 2, 66, 68, 2, 70, 2, 2, 2, 2, 2, 2, 2, 84, 2, 2, 88, 0, 2, 2, 2, 94, 2, 2, 2, 98, 2, 2, 104, 104, 2, 2, 2, 2, 0, 2, 2, 116, 2, 2, 2, 2, 2, 126, 2, 128
Offset: 1

Views

Author

Leroy Quet, Nov 19 2006

Keywords

Comments

From Robert G. Wilson v, Nov 22 2006: (Start)
Only a(1) and a(2) are odd. a(n)=0 for n>1 in A036966.
Only possible values: ..., 0, 1, 2, 10, 16, 24, 26, 32, 36, 46, 62, 66, 68, 70, 84, 88, 94, 98, 104, ..., .
Position of first occurrence: 8, 1, 3, 9, 12, 18, 20, 25, 28, 36, 44, 49, 50, 52, 60, 63, 68, 72, 75, ..., .
(End)

Examples

			12 has a prime factorization of 2^2 *3^1. So a(12) is the sum of the terms among the first 11 terms of the sequence which equal 1 or 2. There are seven 2's and two 1's among the first 11 terms; so a(12) = 1+1+2+2+2+2+2+2+2 = 16.
		

Crossrefs

Cf. A125087.

Programs

  • Mathematica
    f[l_List] := Append[l, Plus @@ Select[l, MemberQ[Last /@ FactorInteger[Length[l] + 1], # ] &]];Nest[f, {1}, 91] (* Ray Chandler, Nov 21 2006 *)
    a[1] = 1; a[n_] := a[n] = Plus @@ Flatten[ Cases[ Array[a, n - 1], # ] & /@ Union@ Last@ Transpose@ FactorInteger@n]; Array[a, 92] (* Robert G. Wilson v, Nov 22 2006 *)
  • PARI
    up_to = 105;
    A125088list(up_to) = { my(v=vector(up_to)); v[1] = 1; for(n=2,up_to,my(es = vecsort(factor(n)[,2],,8)); v[n] = sum(k=1,n-1,v[k]*!!vecsearch(es,v[k]))); (v); };
    v125088 = A125088list(up_to);
    A125088(n) = v125088[n]; \\ Antti Karttunen, Apr 01 2021

Extensions

Extended by Ray Chandler, Nov 21 2006

A135591 a(1)=1; for n > 1, a(n) is number of earlier terms equal to number of proper divisors of n.

Original entry on oeis.org

1, 1, 2, 1, 3, 1, 4, 1, 1, 1, 7, 0, 7, 1, 1, 1, 10, 0, 10, 0, 1, 1, 12, 2, 2, 1, 1, 0, 14, 2, 14, 0, 1, 1, 1, 0, 17, 1, 1, 2, 19, 2, 19, 0, 0, 1, 20, 0, 6, 0, 1, 0, 21, 2, 1, 2, 1, 1, 24, 0, 24, 1, 0, 1, 1, 2, 27, 0, 1, 2, 28, 0, 28, 1, 0, 0, 1, 2, 30, 0, 1, 1, 32, 0, 1, 1, 1, 2, 35, 0, 1, 0, 1, 1, 1, 0, 39
Offset: 1

Views

Author

Katarzyna Matylla, Feb 25 2008

Keywords

Comments

Similar to A125087, but instead of exponents, we use number of proper divisors.

Examples

			a(12)=0 because 12 has 5 proper divisors (1, 2, 3, 4 and 6) and there is no 5 in a(1), a(2), ..., a(11).
		

Crossrefs

Programs

  • Mathematica
    s={1};Do[AppendTo[s,Count[s,DivisorSigma[0,n]-1]],{n,2,97}];s (* James C. McMahon, Apr 16 2025 *)
  • Maxima
    max:1000; f:makelist(0,i,1,max); apr:makelist(0, i, 0, max); f[1]:1; apr[2]:1; for n:2 through max do block(f[n]:apr[divsum(n,0)], apr[f[n]+1]:apr[f[n]+1]+1);

A135592 a(1)=1; for n > 1, a(n) is number of earlier terms equal to number of prime divisors of n.

Original entry on oeis.org

1, 1, 2, 2, 2, 3, 2, 2, 2, 6, 2, 7, 2, 8, 8, 2, 2, 10, 2, 11, 11, 11, 2, 12, 2, 13, 2, 14, 2, 1, 3, 3, 15, 15, 15, 15, 3, 15, 15, 15, 3, 5, 3, 15, 15, 15, 3, 15, 3, 15, 15, 15, 3, 15, 15, 15, 15, 15, 3, 10, 3, 15, 15, 3, 15, 12, 3, 15, 15, 13, 3, 15, 3, 15, 15, 15, 15, 15, 3, 15, 3, 15, 3, 18
Offset: 1

Views

Author

Katarzyna Matylla, Feb 25 2008

Keywords

Comments

Similar to A125087, but instead of exponents, we use number of prime divisors.

Examples

			a(12)=7 because 12 has 2 prime divisors (2 and 3) and there are 7 2's in a(1), a(2), ..., a(11).
		

Crossrefs

Programs

  • Mathematica
    s={1};Do[AppendTo[s,Count[s,PrimeNu[n]]],{n,2,84}];s (* James C. McMahon, Apr 16 2025 *)
  • Maxima
    max:1000; f:makelist(0,i,1,max); apr:makelist(0, i, 1, max); f[1]:1; apr[2]:1; print(1,1); for n:2 through max do block(f[n]:apr[length(ifactors(n))+1], apr[f[n]+1]:apr[f[n]+1]+1);
Showing 1-3 of 3 results.