A006519 -id:A006519 - cp's OEIS frontend

A160348 Minimal recursive sequence such that if a(n) > 0 then always a(n) > a((f(2n+1)-1)/2), where f is defined by f(2n+1) = (3n+2)/A006519(3n+2) for n>=1, that is f(m) = A075677(2*m-1) for odd m.

0, 2, 1, 6, 7, 5, 3, 11, 4, 13, 14, 10, 15, 52, 12, 50, 53, 9, 54, 59, 51, 62, 63, 49, 60, 65, 8, 68, 69, 58, 16, 75, 61, 56, 76, 48, 77, 80, 64, 84, 85, 67, 78, 88, 57, 44

Offset: 0

Vladimir Shevelev, May 10 2009; corrected May 13 2009, May 19 2009

If the (3x+1)-Collatz conjecture is true, then this sequence is a permutation of the nonnegative integers.

			a(0)=0. Let m=3. Then f(m)=5, f^2(m)=1. The corresponding numbers n=(m-1)/2 are 1,2,0. By the condition, a(1) > a(2) > a(0)=0. Therefore let a(2)=1, a(1)=2. Furthermore, consider m=7. Then f(m)=11, f^2(m)=17, f^3(m)=13, f^4(m)=5. The corresponding numbers n=(m-1)/2 are 3,5,8,6,2 and, by the condition, a(3) > a(5) > a(8) > a(6) > a(2)=1. Therefore set a(6)=3 (the minimal value which yet did not appear), a(8)=4, a(5)=5, a(3)=6, etc.

Cf. A006519, A075677, A159885, A159945, A160198, A122458, A259667.

Name edited by Michel Marcus, Feb 01 2021

A162588 G.f.: A(x) = exp( 2Sum_{n>=1} 2^n/A006519(n) x^n/n ), where A006519(n) = highest power of 2 dividing n.

Original entry on oeis.org

1, 4, 10, 24, 52, 112, 240, 512, 1060, 2192, 4552, 9440, 19408, 39872, 81984, 168448, 342632, 696736, 1421200, 2897856, 5891872, 11976064, 24361856, 49543168, 100329952, 203147136, 411939264, 835168512, 1690383744, 3420860928

Offset: 0

Paul D. Hanna, Jul 07 2009

nonn

			G.f.: A(x) = 1 + 4*x + 10*x^2 + 24*x^3 + 52*x^4 + 112*x^5 + 240*x^6 + ...
log(A(x))/2 = 2*x + 2*x^2/2 + 8*x^3/3 + 4*x^4/4 + 32*x^5/5 + 32*x^6/6 + 128*x^7/7 + ...

G. C. Greubel, Table of n, a(n) for n = 0..1000

Cf. A006519, A000123.

nmax = 150; a[n_]:= SeriesCoefficient[Series[Exp[Sum[2^(k + 1 - IntegerExponent[k, 2])*q^k/k, {k, 1, nmax}]], {q, 0, nmax}], n]; Table[a[n], {n,0,50}] (* G. C. Greubel, Jul 04 2018 *)

{a(n)=local(L=2*sum(m=1,n,2^(m-valuation(m,2))*x^m/m)+x*O(x^n));polcoeff(exp(L),n)}

A191769 G.f. A(x) satisfies: A(x) = 1 + Sum_{n>=1} x^n*A(x)^A006519(n) where A006519(n) = highest power of 2 dividing n.

Original entry on oeis.org

1, 1, 2, 5, 12, 33, 92, 267, 792, 2403, 7414, 23199, 73454, 234901, 757654, 2461877, 8051284, 26480681, 87534184, 290652931, 968992200, 3242229475, 10884245838, 36648566551, 123739675390, 418848744517, 1421072269234, 4831811596381

Offset: 0

Paul D. Hanna, Jun 16 2011

nonn

			G.f.: A(x) = 1 + x + 2*x^2 + 5*x^3 + 12*x^4 + 33*x^5 + 92*x^6 + 267*x^7 +...
The g.f. satisfies the following relations:
A(x) = 1 + x*A(x) + x^2*A(x)^2 + x^3*A(x) + x^4*A(x)^4 + x^5*A(x) + x^6*A(x)^2 + x^7*A(x) + x^8*A(x)^8 +...+ x^n*A(x)^A006519(n) +...
A(x) = 1 + x*A(x)/(1-x^2) + x^2*A(x)^2/(1-x^4) + x^4*A(x)^4/(1-x^8) + x^8*A(x)^8/(1-x^16) + x^16*A(x)^16/(1-x^32) +...

Cf. A191768.

{a(n)=local(A=1+x);for(i=1,n,A=1+sum(m=1,n,x^m*(A+x*O(x^n))^(2^valuation(m,2))));polcoeff(A,n)}

G.f. A(x) satisfies: A(x) = 1 + Sum_{n>=0} x^(2^n)*A(x)^(2^n)/(1 - x^(2^(n+1))).

A233808 Autosequence preceding A198631(n)/A006519(n+1). Numerators.

Original entry on oeis.org

0, 0, 1, 3, 3, 5, 5, 7, 7, -3, -3, 121, 121, -1261, -1261, 20583, 20583, -888403, -888403, 24729925, 24729925, -862992399, -862992399, 36913939769, 36913939769, -1899853421885, -1899853421885

Offset: 0

Paul Curtz, Dec 16 2013

The fractions are g(n)=0, 0, 1, 3/2, 3/2, 5/4, 5/4, 7/4, 7/4, -3/8, -3/8, 121/8, 121/8, -1261/8, -1261/8, 20583/8, 20583/8, -888403/16, -888403/16,... . The denominators are 1, 1, followed by A053644(n+1).
g(n+2) - g(n+1) = A198631(n)/A006519(n+1).
The corresponding fractions to g(n) are f(n) in A165142(n).
g(n) differences table:
0,       0,    1,  3/2,  3/2,  5/4,
0,       1,  1/2,    0, -1/4,    0,
1,    -1/2, -1/2, -1/4,  1/4,  1/2,    Euler twin numbers (new),
-3/2,    0,  1/4,  1/2,  1/4,   -1,
3/2,   1/4,  1/4, -1/4, -5/4, -5/8,
-5/4,    0, -1/2,   -1,  5/8, 13/2, etc.
Like A198631(n)/A006519(n+1),g(n) is an autosequence of the second kind.
If we proceed, here for Euler polynomials, like in A233565 for Bernoulli polynomials, we obtain
1) A133138(n)/A007395(n) (unreduced form) or
2) A233508(n)/A232628(n) (reduced form),the first array in A133135.
The Bernoulli's corresponding fractions to 1) are A193815(n)/(A003056(n) with 1 instead of 0).

Cf. A051716/A051717, Bernoulli twin numbers.

max = 27; p[0] = 1; p[n_] := (1 + x)*((1 + x)^(n - 1) + x^(n - 1))/2; t = Table[Coefficient[p[n], x, k], {n, 0, max + 2}, {k, 0, max + 2}]; a[n_] := (-1)^n*Inverse[t][[n, 2]] // Numerator; a[0] = 0; Table[a[n], {n, 0, max}] (* Jean-François Alcover, Jan 11 2016 *)

a(n) = 0, 0, followed by (-1)^n *A141424(n).

A349345 Sum of A109168 and its Dirichlet inverse, where A109168(n) = (n+A006519(n))/2, and A006519 is the highest power of 2 dividing n.

Original entry on oeis.org

2, 0, 0, 4, 0, 8, 0, 8, 4, 12, 0, 8, 0, 16, 12, 16, 0, 12, 0, 12, 16, 24, 0, 16, 9, 28, 12, 16, 0, 8, 0, 32, 24, 36, 24, 20, 0, 40, 28, 24, 0, 12, 0, 24, 26, 48, 0, 32, 16, 34, 36, 28, 0, 32, 36, 32, 40, 60, 0, 32, 0, 64, 36, 64, 42, 20, 0, 36, 48, 24, 0, 40, 0, 76, 46, 40, 48, 24, 0, 48, 37, 84, 0, 44, 54, 88, 60, 48

Offset: 1

Antti Karttunen, Nov 15 2021

Antti Karttunen, Table of n, a(n) for n = 1..20000

Cf. A109168, A349344.

up_to = 20000;
DirInverseCorrect(v) = { my(u=vector(#v)); u[1] = (1/v[1]); for(n=2, #v, u[n] = (-u[1]*sumdiv(n, d, if(dA109168(n) = ((n+bitand(n, -n))\2); \\ From A109168
v349344 = DirInverseCorrect(vector(up_to,n,A109168(n)));
A349344(n) = v349344[n];
A349345(n) = (A109168(n)+A349344(n));

a(n) = A109168(n) + A349344(n).
a(1) = 2, and for n > 1, a(n) = -Sum_{d|n, 1A109168(d) * A349344(n/d).
For all n >= 1, a(4*n) = 4*A109168(n). - Antti Karttunen, Dec 07 2021

A162583 G.f.: A(x) = exp( 6Sum_{n>=1} A006519(n)A038500(n) * x^n/n ).

Original entry on oeis.org

1, 6, 24, 78, 222, 570, 1356, 3030, 6432, 13074, 25608, 48558, 89502, 160854, 282624, 486534, 822174, 1365978, 2234400, 3602742, 5732202, 9008034, 13993320, 21503730, 32711460, 49287750, 73598280, 108968334, 160041750, 233262786

Offset: 0

Paul D. Hanna, Jul 06 2009

nonn

A006519(n) = highest power of 2 dividing n and

A038500(n) = highest power of 3 dividing n.

			G.f.: A(x) = 1 + 6*x + 24*x^2 + 78*x^3 + 222*x^4 + 570*x^5 + 1356*x^6 + ...
log(A(x))/6 = x + 2*x^2/2 + 3*x^3/3 + 4*x^4/4 + x^5/5 + 6*x^6/6 + x^7/7 + 8*x^8/8 + 9*x^9/9 + ... + A006519(n)*A038500(n)*x^n/n + ...

G. C. Greubel, Table of n, a(n) for n = 0..1000

Cf. A006519, A038500.

nmax = 150; a[n_]:= SeriesCoefficient[Series[Exp[Sum[2^(IntegerExponent[k, 2] + 1)*3^(IntegerExponent[k, 3] + 1)*q^k/k, {k, 1, nmax}]], {q, 0, nmax}], n]; Table[a[n], {n, 0, 50}] (* G. C. Greubel, Jul 04 2018 *)

{a(n)=local(L=sum(m=1,n,6*2^valuation(m,2)*3^valuation(m,3)*x^m/m)+x*O(x^n));polcoeff(exp(L),n)}

A162585 G.f.: A(x) = exp( Sum_{n>=1} C(2n,n)A006519(n) x^n/n ), where A006519(n) = highest power of 2 dividing n.

Original entry on oeis.org

1, 2, 8, 20, 114, 288, 1156, 3256, 23464, 59716, 243212, 699216, 3659988, 10265800, 42353168, 128163440, 1127515970, 2858004752, 11768578868, 34294832344, 180335471424, 513911386232, 2137413847256, 6572758142016, 41948816796852

Offset: 0

Paul D. Hanna, Jul 06 2009

nonn

Compare g.f. to the g.f. of the Catalan numbers: exp( Sum_{n>=1} C(2n,n)*x^n/n ), where C(2n,n) form the central binomial coefficients (A000984).

			G.f.: A(x) = 1 + 2*x + 6*x^2 + 10*x^3 + 146*x^4 + 282*x^5 + 826*x^6 + ...
log(A(x)) = 2*x + 12*x^2/2 + 20*x^3/3 + 280*x^4/4 + 252*x^5/5 + 1848*x^6/6 + ... + C(2n,n)*A006519(n)*x^n/n + ...

G. C. Greubel, Table of n, a(n) for n = 0..1000

Cf. A000108, A000123, A000984, A006519.

nmax=50; CoefficientList[Series[Exp[Sum[2^(IntegerExponent[k, 2])*Binomial[2*k, k]*q^k/k, {k,nmax+3}]], {q,0,nmax}], q]  (* G. C. Greubel, Jul 04 2018 *)

{a(n)=local(L=sum(m=1,n,2^valuation(m,2)*binomial(2*m,m)*x^m/m)+x*O(x^n));polcoeff(exp(L),n)}

A162589 G.f.: A(x) = exp( Sum_{n>=1} 2^nA006519(n) x^n/n ), where A006519(n) = highest power of 2 dividing n.

Original entry on oeis.org

1, 2, 6, 12, 38, 76, 188, 376, 1094, 2188, 5236, 10472, 26076, 52152, 118840, 237680, 612678, 1225356, 2804420, 5608840, 13279604, 26559208, 59074504, 118149008, 277925148, 555850296, 1228260104, 2456520208, 5552652792, 11105305584

Offset: 0

Paul D. Hanna, Jul 07 2009

nonn

			G.f.: A(x) = 1 + 2*x + 6*x^2 + 12*x^3 + 38*x^4 + 76*x^5 + 188*x^6 + ...
log(A(x)) = 2*x + 8*x^2/2 + 8*x^3/3 + 64*x^4/4 + 32*x^5/5 + 128*x^6/6 + 128*x^7/7 + ...

G. C. Greubel, Table of n, a(n) for n = 0..1000

Cf. A162588, A006519, A000123.

nmax = 150; a[n_]:= SeriesCoefficient[Series[Exp[Sum[2^(k + IntegerExponent[k, 2])*q^k/k, {k, 1, nmax}]], {q,0,nmax}], n]; Table[a[n], {n,0,50}] (* G. C. Greubel, Jul 04 2018 *)

{a(n)=local(L=sum(m=1,n,2^(m+valuation(m,2))*x^m/m)+x*O(x^n));polcoeff(exp(L),n)}

A174238 Inverse Moebius transform of even part of n (A006519).

Original entry on oeis.org

1, 3, 2, 7, 2, 6, 2, 15, 3, 6, 2, 14, 2, 6, 4, 31, 2, 9, 2, 14, 4, 6, 2, 30, 3, 6, 4, 14, 2, 12, 2, 63, 4, 6, 4, 21, 2, 6, 4, 30, 2, 12, 2, 14, 6, 6, 2, 62, 3, 9, 4, 14, 2, 12, 4, 30, 4, 6, 2, 28, 2, 6, 6, 127, 4, 12, 2, 14, 4, 12, 2, 45, 2, 6, 6, 14, 4, 12, 2, 62

Offset: 1

Ralf Stephan, Nov 27 2010

The Dirichlet g.f. is the Dirichlet g.f. of A006519 multiplied by zeta(s). - R. J. Mathar, Feb 06 2011

Multiplicative because A006519 is. - Andrew Howroyd, Jul 27 2018

Amiram Eldar, Table of n, a(n) for n = 1..10000 (terms 1..1000 from Andrew Howroyd)

Cf. A000005, A006519, A001227.

a[n_] := Sum[2^IntegerExponent[d, 2], {d, Divisors[n]}];
Array[a, 80] (* Jean-François Alcover, Feb 16 2020, from PARI *)
f[p_, e_] := If[p==2, 2^(e+1)-1, e+1]; a[1] = 1; a[n_] := Times @@ (f @@@ FactorInteger[n]); Array[a, 100] (* Amiram Eldar, Sep 30 2020 *)

a(n) = sumdiv(n, d, 2^valuation(d, 2)); \\ Michel Marcus, Mar 27 2015

a(1) = 1, a(2n) = 2a(n) + A001227(n), a(2n+1) = A000005(2n+1).
Dirichlet g.f.:  zeta(s)^2*(1-2^(-s))/(1-2^(-s+1)). - Ralf Stephan, Mar 27 2015
Multiplicative with a(2^e) = 2^(e+1)-1, and a(p^e) = e+1 for p > 2. - Amiram Eldar, Sep 30 2020
Sum_{k=1..n} a(k) ~ n*(log(n)^2/(4*log(2)) + (3/4 - 1/(2*log(2)) + gamma/log(2))*log(n) - 3/4 + log(2)/24 + 1/(2*log(2)) + (3/2 - 1/log(2))*gamma + gamma^2/(2*log(2)) - sg1/log(2)), where gamma is the Euler-Mascheroni constant A001620 and sg1 is the first Stieltjes constant (see A082633). - Vaclav Kotesovec, Nov 20 2021

Title corrected by R. J. Mathar, Feb 06 2011

Terms a(61) and beyond from Andrew Howroyd, Jul 27 2018

A237425 Denominators of A164555(n)/A027642(n) + A198631(n)/A006519(n+1).

Original entry on oeis.org

1, 1, 6, 4, 30, 2, 42, 8, 30, 2, 66, 4, 2730, 2, 6, 16, 510, 2, 798, 4, 330, 2, 138, 8, 2730, 2, 6, 4, 870, 2, 14322, 32, 510, 2, 6, 4, 1919190, 2, 6, 8, 13530, 2, 1806, 4, 690, 2, 282, 16, 46410, 2, 66, 4, 1590, 2, 798, 8

Offset: 0

Paul Curtz, Feb 07 2014

nonn

An autosequence is a sequence which has its inverse binomial transform equal to the signed sequence. There are two possibilities. For the first kind, the main diagonal is 0's=A000004, the first two following diagonals being the same (generally not A000004). Integers example: A000045(n).
For the second kind, the main diagonal is the double of the following diagonal. Example: the companion to A000045(n) is A000032(n)=2, 1, 3, ... .
A000032(n)/2 is also a possibility. Here a(n) is the denominator of the sum of two autosequences of second kind involving (fractional) Euler and Bernoulli numbers. The corresponding fractional sequence is also an autosequence of the second kind: 2, 1, 1/6, -1/4, -1/30, 1/2, 1/42, -17/8, -1/30, 31/2, 5/66, -691/4, -691/2730,... . It could be divided by 2.

Cf. 1/n in A003506, A194767, A218289, A168516/A168426, A224270/A046161.

a[n_] := BernoulliB[n] + EulerE[n, 1]/2^IntegerExponent[n, 2]; a[0] = 2; a[1] = 1; Table[a[n] // Denominator, {n, 0, 55}] (* Jean-François Alcover, Feb 11 2014 *)

a(2n) = A002445(n). a(2n+2) = A171977(n+2).

cp's OEIS Frontend

A160348 Minimal recursive sequence such that if a(n) > 0 then always a(n) > a((f(2n+1)-1)/2), where f is defined by f(2n+1) = (3n+2)/A006519(3n+2) for n>=1, that is f(m) = A075677(2*m-1) for odd m.

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A162588 G.f.: A(x) = exp( 2*Sum_{n>=1} 2^n/A006519(n) * x^n/n ), where A006519(n) = highest power of 2 dividing n.

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A191769 G.f. A(x) satisfies: A(x) = 1 + Sum_{n>=1} x^n*A(x)^A006519(n) where A006519(n) = highest power of 2 dividing n.

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A233808 Autosequence preceding A198631(n)/A006519(n+1). Numerators.

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A349345 Sum of A109168 and its Dirichlet inverse, where A109168(n) = (n+A006519(n))/2, and A006519 is the highest power of 2 dividing n.

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A162583 G.f.: A(x) = exp( 6*Sum_{n>=1} A006519(n)*A038500(n) * x^n/n ).

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A162585 G.f.: A(x) = exp( Sum_{n>=1} C(2n,n)*A006519(n) * x^n/n ), where A006519(n) = highest power of 2 dividing n.

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A162589 G.f.: A(x) = exp( Sum_{n>=1} 2^n*A006519(n) * x^n/n ), where A006519(n) = highest power of 2 dividing n.

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A174238 Inverse Moebius transform of even part of n (A006519).

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A162588 G.f.: A(x) = exp( 2Sum_{n>=1} 2^n/A006519(n) x^n/n ), where A006519(n) = highest power of 2 dividing n.

A162583 G.f.: A(x) = exp( 6Sum_{n>=1} A006519(n)A038500(n) * x^n/n ).

A162585 G.f.: A(x) = exp( Sum_{n>=1} C(2n,n)A006519(n) x^n/n ), where A006519(n) = highest power of 2 dividing n.

A162589 G.f.: A(x) = exp( Sum_{n>=1} 2^nA006519(n) x^n/n ), where A006519(n) = highest power of 2 dividing n.