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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A197953 a(n) = 1 + Sum_{d|n, d>1} d * a(n/d).

Original entry on oeis.org

1, 3, 4, 11, 6, 24, 8, 43, 22, 38, 12, 128, 14, 52, 54, 171, 18, 186, 20, 206, 74, 80, 24, 640, 56, 94, 130, 284, 30, 494, 32, 683, 114, 122, 118, 1226, 38, 136, 134, 1038, 42, 682, 44, 440, 432, 164, 48, 3072, 106, 488, 174, 518, 54, 1374, 182, 1436, 194
Offset: 1

Views

Author

Paul D. Hanna, Oct 19 2011

Keywords

Comments

Logarithmic derivative of A129374, where g.f. G(x) of A129374 satisfies: G(x) = 1/(1-x) * G(x^2)*G(x^3)*G(x^4)*...*G(x^n)*...

Examples

			L.g.f.: L(x) = x + 3*x^2/2 + 4*x^3/3 + 11*x^4/4 + 6*x^5/5 + 24*x^6/6 +...
where
L(x) = -log(1-x) + L(x^2) + L(x^3) + L(x^4) + L(x^5) +...+ L(x^n) +...
also, exp(L(x)) is the g.f. of A129374:
exp(L(x)) = 1 + x + 2*x^2 + 3*x^3 + 6*x^4 + 8*x^5 + 15*x^6 + 20*x^7 +...
		

Crossrefs

Programs

  • PARI
    {a(n)=sumdiv(n,d,d*if(d==1,1,a(n/d)))}
    
  • PARI
    /* L.g.f. satisfies: L(x) = -log(1-x) + Sum_{n>1} L(x^n) */
    {a(n)=local(L=x,X=x+x*O(x^n));for(i=1,n,L=-log(1-X)+sum(m=2,n,subst(L,x,x^m+x*O(x^n))));n*polcoeff(L,n)}

Formula

L.g.f. satisfies: L(x) = -log(1-x) + Sum_{n>1} L(x^n), where L(x) = Sum_{n>=1} a(n)*x^n/n.

A317580 Number of unlabeled rooted identity trees with n nodes and a distinguished leaf.

Original entry on oeis.org

1, 1, 1, 3, 5, 12, 28, 66, 153, 367, 880, 2121, 5127, 12441, 30248, 73746, 180077, 440571, 1079438, 2648511, 6506170, 16001256, 39393173, 97074140, 239419963, 590972968, 1459808862, 3608483107, 8925476591, 22090139751, 54702648393, 135533335933, 335967782916
Offset: 1

Views

Author

Gus Wiseman, Jul 31 2018

Keywords

Comments

Total number of leaves in all rooted identity trees with n nodes. - Andrew Howroyd, Aug 28 2018

Examples

			The a(6) = 12 rooted identity trees with a distinguished leaf:
(((((O))))),
(((O(o)))), (((o(O)))),
((O((o)))), ((o((O)))),
(O(((o)))), (o(((O)))),
((O)((o))), ((o)((O))),
(O(o(o))), (o(O(o))), (o(o(O))).
		

Crossrefs

Programs

  • Mathematica
    urit[n_]:=Join@@Table[Select[Union[Sort/@Tuples[urit/@ptn]],UnsameQ@@#&],{ptn,IntegerPartitions[n-1]}];
    Table[Sum[Length[Flatten[{t/.{}->1}]],{t,urit[n]}],{n,10}]
  • PARI
    WeighMT(u)={my(n=#u, p=x*Ser(u), vars=variables(p)); Vec(exp( sum(i=1, n, (-1)^(i-1)*substvec(p + O(x*x^(n\i)), vars, apply(v->v^i,vars))/i ))-1)}
    seq(n)={my(v=[y]); for(n=2, n, v=concat([y], WeighMT(v))); apply(p -> subst(deriv(p), y, 1), v)} \\ Andrew Howroyd, Aug 28 2018

Formula

a(n) = Sum_{k=1, n} k*A055327(n, k). - Andrew Howroyd, Aug 28 2018

Extensions

Terms a(26) and beyond from Andrew Howroyd, Aug 28 2018

A335062 a(n) = 1 - Sum_{d|n, d > 1} (-1)^d * a(n/d).

Original entry on oeis.org

1, 0, 2, 0, 2, -2, 2, 0, 4, -2, 2, 0, 2, -2, 6, 0, 2, -8, 2, 0, 6, -2, 2, 0, 4, -2, 8, 0, 2, -14, 2, 0, 6, -2, 6, 4, 2, -2, 6, 0, 2, -14, 2, 0, 16, -2, 2, 0, 4, -8, 6, 0, 2, -24, 6, 0, 6, -2, 2, 8, 2, -2, 16, 0, 6, -14, 2, 0, 6, -14, 2, 0, 2, -2, 16, 0, 6, -14, 2, 0, 16
Offset: 1

Views

Author

Ilya Gutkovskiy, May 21 2020

Keywords

Comments

Inverse Moebius transform of A308077.

Crossrefs

Cf. A048298, A065091 (positions of 2's), A067824, A067856, A308077, A325144, A335283.

Programs

  • Mathematica
    a[n_] := a[n] = 1 - DivisorSum[n, (-1)^# a[n/#] &, # > 1 &]; Table[a[n], {n, 1, 81}]
  • PARI
    lista(nn) = {my(va = vector(nn)); for (n=1, nn, va[n] = 1 - sumdiv(n, d, if (d>1, (-1)^d*va[n/d]));); va;} \\ Michel Marcus, May 22 2020

Formula

G.f. A(x) satisfies: A(x) = x / (1 - x) - Sum_{k>=2} (-1)^k * A(x^k).

A337104 Number of strict chains of divisors from n! to 1 using terms of A130091 (numbers with distinct prime multiplicities).

Original entry on oeis.org

1, 1, 1, 0, 14, 0, 384, 0, 0, 0, 21077680, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0
Offset: 0

Views

Author

Gus Wiseman, Aug 17 2020

Keywords

Comments

The support appears to be {0, 1, 2, 4, 6, 10}.

Examples

			The a(4) = 14 chains:
  24/1
  24/2/1
  24/3/1
  24/4/1
  24/8/1
  24/12/1
  24/4/2/1
  24/8/2/1
  24/8/4/1
  24/12/2/1
  24/12/3/1
  24/12/4/1
  24/8/4/2/1
  24/12/4/2/1
		

Crossrefs

A336867 appears to be the positions of zeros.
A336868 is the characteristic function (image under A057427).
A336942 is the version for superprimorials (n > 1).
A337105 does not require distinct prime multiplicities.
A337074 does not require chains to end with 1.
A337075 is the version for chains not containing n!.
A000005 counts divisors.
A000142 lists factorial numbers.
A001055 counts factorizations.
A027423 counts divisors of factorial numbers.
A067824 counts chains of divisors starting with n.
A074206 counts chains of divisors from n to 1.
A130091 lists numbers with distinct prime multiplicities.
A181796 counts divisors with distinct prime multiplicities.
A253249 counts chains of divisors.
A327498 gives the maximum divisor with distinct prime multiplicities.
A336414 counts divisors of n! with distinct prime multiplicities.
A336423 counts chains using A130091, with maximal case A336569.
A336425 counts divisible pairs of divisors of n!, both in A130091.
A336571 counts chains of divisors 1 < d < n using A130091.
A337071 counts chains of divisors starting with n!.

Programs

  • Mathematica
    strchns[n_]:=If[n==1,1,If[!UnsameQ@@Last/@FactorInteger[n],0,Sum[strchns[d],{d,Select[DeleteCases[Divisors[n],n],UnsameQ@@Last/@FactorInteger[#]&]}]]];
    Table[strchns[n!],{n,0,8}]

Formula

a(n) = A337075(n) whenever A337075(n) != 0.
a(n) = A337074(n)/2 for n > 1.
a(n) = A336423(n!).

A343390 a(n) = 1 + Product_{d|n, d < n} a(d).

Original entry on oeis.org

2, 3, 3, 7, 3, 19, 3, 43, 7, 19, 3, 2395, 3, 19, 19, 1807, 3, 2395, 3, 2395, 19, 19, 3, 246546091, 7, 19, 43, 2395, 3, 370387, 3, 3263443, 19, 19, 19, 96124306951, 3, 19, 19, 246546091, 3, 370387, 3, 2395, 2395, 19, 3, 109838449356687381331, 7, 2395, 19, 2395, 3, 246546091, 19
Offset: 1

Views

Author

Ilya Gutkovskiy, Apr 13 2021

Keywords

Crossrefs

Programs

  • Maple
    a:= proc(n) option remember;
          1+mul(a(d), d=numtheory[divisors](n) minus {n})
        end:
    seq(a(n), n=1..55);  # Alois P. Heinz, Apr 13 2021
  • Mathematica
    a[n_] := a[n] = 1 + Product[If[d < n, a[d], 1], {d, Divisors[n]}]; Table[a[n], {n, 55}]

Formula

a(p^k) = A000058(k) for p prime.

A306921 Number of ways of breaking the binary expansion of n into consecutive blocks with no leading zeros.

Original entry on oeis.org

1, 1, 2, 2, 3, 3, 4, 4, 4, 4, 6, 6, 6, 6, 8, 8, 5, 5, 8, 8, 9, 9, 12, 12, 8, 8, 12, 12, 12, 12, 16, 16, 6, 6, 10, 10, 12, 12, 16, 16, 12, 12, 18, 18, 18, 18, 24, 24, 10, 10, 16, 16, 18, 18, 24, 24, 16, 16, 24, 24, 24, 24, 32, 32, 7, 7, 12, 12, 15, 15, 20, 20, 16
Offset: 0

Views

Author

Peter Kagey, Mar 16 2019

Keywords

Comments

The number 0 is not considered to have a leading zero.
a(n) >= 2^(A000120(n) - 1).
a(2^n - 1) = 2^(n-1) for n > 0.
a(2^n) = n+1.
Conjecture: n appears A067824(n) times for n > 1.

Examples

			For n = 13 the a(13) = 6 partitions are [1101], [1, 101], [110, 1], [1, 10, 1], [11, 0, 1], and [1, 1, 0, 1].
Notice that [1, 1, 01] and [11, 01] are not valid partitions because the last part has a leading zero.
		

Crossrefs

A321318 gives number of distinct sums of such partitions.

Formula

From Charlie Neder, May 08 2019: (Start)
If n = k*2^e + {0,1} with k odd and e > 0, then a(n) = a(k)*(e+1).
Proof: Each partition of n is uniquely determined by a partition of k (call it K) and a choice of some number, from 0 to e, of trailing digits to append to the final part in K, since any remaining digits must appear as singletons. The conjecture follows, since each ordered factorization of a number m produces two numbers n such that a(n) = m, one of each parity, and A067824(n) = 2*A074206(n).
Corollary: For n >= 1, a(2n) = a(2n+1) = Product{k+1 | k in row n of A066099}. (End)

A316789 Number of same-tree-factorizations of n.

Original entry on oeis.org

1, 1, 1, 2, 1, 1, 1, 2, 2, 1, 1, 1, 1, 1, 1, 6, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 2, 1, 1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 14, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 6, 1, 1, 1, 1, 1, 1
Offset: 1

Views

Author

Gus Wiseman, Jul 14 2018

Keywords

Comments

A constant factorization of n is a finite nonempty constant multiset of positive integers greater than 1 with product n. Constant factorizations correspond to perfect divisors (A089723). A same-tree-factorization of n is either (case 1) the number n itself or (case 2) a finite sequence of two or more same-tree-factorizations, one of each factor in a constant factorization of n.
a(n) depends only on the prime signature of n. - Andrew Howroyd, Nov 18 2018

Examples

			The a(64) = 14 same-tree-factorizations:
  64
  (8*8)
  (4*4*4)
  (8*(2*2*2))
  ((2*2*2)*8)
  (4*4*(2*2))
  (4*(2*2)*4)
  ((2*2)*4*4)
  (2*2*2*2*2*2)
  (4*(2*2)*(2*2))
  ((2*2)*4*(2*2))
  ((2*2)*(2*2)*4)
  ((2*2*2)*(2*2*2))
  ((2*2)*(2*2)*(2*2))
		

Crossrefs

Programs

  • Mathematica
    a[n_]:=1+Sum[a[n^(1/d)]^d,{d,Rest[Divisors[GCD@@FactorInteger[n][[All,2]]]]}]
    Array[a,100]
  • PARI
    a(n)={my(z, e=ispower(n,,&z)); 1 + if(e, sumdiv(e, d, if(d>1, a(z^(e/d))^d)))} \\ Andrew Howroyd, Nov 18 2018

Formula

a(n) = 1 + Sum_{n = x^y, y > 1} a(x)^y.
a(2^n) = A281145(n).

A316790 Number of orderless same-tree-factorizations of n.

Original entry on oeis.org

1, 1, 1, 2, 1, 1, 1, 2, 2, 1, 1, 1, 1, 1, 1, 5, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 2, 1, 1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 9, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 5, 1, 1, 1, 1, 1, 1
Offset: 1

Views

Author

Gus Wiseman, Jul 14 2018

Keywords

Comments

A constant factorization of n is a finite nonempty constant multiset of positive integers greater than 1 with product n. Constant factorizations correspond to perfect divisors (A089723). An orderless same-tree-factorization of n is either (case 1) the number n itself or (case 2) a finite multiset of two or more orderless same-tree-factorizations, one of each factor in a constant factorization of n.
a(n) depends only on the prime signature of n. - Andrew Howroyd, Nov 18 2018

Examples

			The a(64) = 9 orderless same-tree-factorizations:
  64
  (8*8)
  (4*4*4)
  (4*4*(2*2))
  (8*(2*2*2))
  (2*2*2*2*2*2)
  (4*(2*2)*(2*2))
  ((2*2*2)*(2*2*2))
  ((2*2)*(2*2)*(2*2))
		

Crossrefs

Programs

  • Mathematica
    a[n_]:=1+Sum[Binomial[a[n^(1/d)]+d-1,d],{d,Rest[Divisors[GCD@@FactorInteger[n][[All,2]]]]}]
    Array[a,100]
  • PARI
    a(n)={my(z, e=ispower(n,,&z)); 1 + if(e, sumdiv(e, d, if(d>1, binomial(a(z^(e/d)) + d - 1, d))))} \\ Andrew Howroyd, Nov 18 2018

Formula

a(n) = 1 + Sum_{n = x^y, y > 1} binomial(a(x) + y - 1, y).
a(2^n) = A289078(n).

A317581 a(1) = 1; a(n > 1) = 1 + Sum_{d|n, d

Original entry on oeis.org

1, 0, 0, 1, 0, 2, 0, 0, 1, 2, 0, -2, 0, 2, 2, 1, 0, -2, 0, -2, 2, 2, 0, 4, 1, 2, 0, -2, 0, -6, 0, 0, 2, 2, 2, 7, 0, 2, 2, 4, 0, -6, 0, -2, -2, 2, 0, -4, 1, -2, 2, -2, 0, 4, 2, 4, 2, 2, 0, 16, 0, 2, -2, 1, 2, -6, 0, -2, 2, -6, 0, -12, 0, 2, -2, -2, 2, -6, 0, -4
Offset: 1

Views

Author

Gus Wiseman, Jul 31 2018

Keywords

Comments

If p is prime, a(p^k) = 0 if k is odd, 1 if k is even. - Robert Israel, Aug 01 2018

Crossrefs

Programs

  • Maple
    f:= n -> 1 + add(numtheory:-mobius(n/d)*procname(d),d=numtheory:-divisors(n) minus {n}):
    f(1):= 1:
    map(f, [$1..100]); # Robert Israel, Aug 01 2018
  • Mathematica
    a[n_]:=1+Sum[MoebiusMu[n/d]*a[d],{d,Most[Divisors[n]]}];
    Array[a,100]
  • Python
    from sympy import mobius, divisors
    def A317581(n): return 1 + (0 if n == 1 else sum(mobius(n//d)*A317581(d) for d in divisors(n,generator=True) if d < n)) # Chai Wah Wu, Jan 14 2022

A335283 a(n) = 1 + Sum_{d|n, n/d odd, d < n} a(d).

Original entry on oeis.org

1, 1, 2, 1, 2, 2, 2, 1, 4, 2, 2, 2, 2, 2, 6, 1, 2, 4, 2, 2, 6, 2, 2, 2, 4, 2, 8, 2, 2, 6, 2, 1, 6, 2, 6, 4, 2, 2, 6, 2, 2, 6, 2, 2, 16, 2, 2, 2, 4, 4, 6, 2, 2, 8, 6, 2, 6, 2, 2, 6, 2, 2, 16, 1, 6, 6, 2, 2, 6, 6, 2, 4, 2, 2, 16, 2, 6, 6, 2, 2, 16, 2, 2, 6, 6, 2, 6, 2, 2, 16, 6, 2, 6, 2, 6, 2, 2, 4, 16, 4, 2, 6, 2, 2, 26
Offset: 1

Views

Author

Ilya Gutkovskiy, May 30 2020

Keywords

Crossrefs

Cf. A000079 (positions of 1's), A038550 (positions of 2's), A067824, A074206, A209229, A307779, A335062.

Programs

  • Mathematica
    a[n_] := 1 + Sum[If[d < n && OddQ[n/d], a[d], 0], {d, Divisors[n]}]; Table[a[n], {n, 85}]
    terms = 85; A[] = 0; Do[A[x] = x/(1 - x) + Sum[A[x^(2 k - 1)], {k, 2, terms}] + O[x]^(terms + 1) // Normal, terms + 1]; CoefficientList[A[x], x] // Rest
  • PARI
    up_to = 20000;
    A335283lista(up_to) = {my(v = vector(up_to)); for(n=1, up_to, v[n] = 1 + sumdiv(n, d, if(dA335283lista(up_to);
    A335283(n) = v335283[n]; \\ Antti Karttunen, Dec 09 2021

Formula

G.f. A(x) satisfies: A(x) = x / (1 - x) + Sum_{k>=2} A(x^(2*k-1)).
G.f.: x / (1 - x) + Sum_{n>=1} a(n) * x^(3*n) / (1 - x^(2*n)).
a(1) = 1; a(2*n) = a(n), a(2*n+1) = 2 * A074206(2*n+1).

Extensions

More terms from Antti Karttunen, Dec 09 2021
Previous Showing 81-90 of 98 results. Next