A010036 -id:A010036 - cp's OEIS frontend

A125185 Triangle read by rows: T(n,k) is the coefficient of t^k in the polynomial S(n,t)=[(1+t)(2+t)^n+(1-t)t^n]/2 (0<=k<=n).

1, 1, 2, 2, 4, 3, 4, 10, 9, 4, 8, 24, 28, 16, 5, 16, 56, 80, 60, 25, 6, 32, 128, 216, 200, 110, 36, 7, 64, 288, 560, 616, 420, 182, 49, 8, 128, 640, 1408, 1792, 1456, 784, 280, 64, 9, 256, 1408, 3456, 4992, 4704, 3024, 1344, 408, 81, 10, 512, 3072, 8320, 13440, 14400

Offset: 0

Emeric Deutsch, Dec 04 2006

Sum of terms in row n = 3^n. The polynomials S(n,t) can be defined recursively by S(0,t)=1, S(n,t)=t^n - t^(n-1) + (2+t)S(n-1,t) for n>=1. S(n,t)=Sum(B(j,t), j=2^n .. 2^(n+1)-1), where B(n,t) are the Stern polynomials, defined by B(0,t)=0, B(1,t)=1, B(2n,t)=tB(n,t), B(2n+1,t)=B(n+1,t)+B(n,t) for n>=1 (see S. Klavzar et al. and A125184). For example, S(2,t)=B(4,t)+B(5,t)+B(6,t)+B(7,t).

Subtriangle of (0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, ...) DELTA (1, 1, -1, 1, 0, 0, 0, 0, 0, 0, 0, 0, ...) where DELTA is the operator defined in A084938. - Philippe Deléham, Feb 26 2012

			Triangle starts:
  1;
  1,2;
  2,4,3;
  4,10,9,4;
  8,24,28,16,5;
  16,56,80,60,25,6;
Triangle (0,1,1,0,0,0,...) DELTA (1,1,-1,1,0,0,0,0,...) begins:
  1
  0, 1
  0, 1, 2
  0, 2, 4, 3
  0, 4, 10, 9, 4
  0, 8, 24, 28, 16, 5
  0, 16, 56, 80, 60, 25, 6

S. Klavzar, U. Milutinovic and C. Petr, Stern polynomials, Adv. Appl. Math. 39 (2007), 86-95.

Cf. A125184.

Maple
```
T:=proc(n,k) if k
				
```

T(n,k)=2^(n-k-1)*(n+k+1)binomial(n,k)/(n-k+1) if k

G.f.: (1-x)/((1-y*x)*(1-(y+2)*x)). - Philippe Deléham, Feb 26 2012

Recurrence : T(n,k) = 2*T(n-1,k) + 2*T(n-1,k-1) - 2*T(n-2,k-1) - T(n-2,k-2) with T(0,0) = T(1,0) = 1, T(n,k) = 0 if k<0 or if k>n. - Philippe Deléham, Feb 26 2012

Sum_{k=0..n} T(n,k)*x^k = A033999(n), A011782(n), A000244(n), A010036(n), A081625(n) for x = -1, 0, 1, 2, 3 respectively. - Philippe Deléham, Feb 26 2012

A167993 Expansion of x^2/((3x-1)(3*x^2-1)).

Original entry on oeis.org

0, 0, 1, 3, 12, 36, 117, 351, 1080, 3240, 9801, 29403, 88452, 265356, 796797, 2390391, 7173360, 21520080, 64566801, 193700403, 581120892, 1743362676, 5230147077, 15690441231, 47071500840, 141214502520, 423644039001, 1270932117003, 3812797945332, 11438393835996

Offset: 0

Paul Curtz, Nov 16 2009

The terms satisfy a(n) = 3*a(n-1) +3*a(n-2) -9*a(n-3), so they follow the pattern a(n) = p*a(n-1) +q*a(n-2) -p*q*a(n-3) with p=q=3. This could be called the principal sequence for that recurrence because we have set all but one of the initial terms to zero. [p=q=1 leads to the principal sequence A004526. p=q=2 leads essentially to A032085. The common feature is that the denominator of the generating function does not have a root at x=1, so the sequences of higher order successive differences have the same recurrence as the original sequence. See A135094, A010036, A006516.]

Vincenzo Librandi, Table of n, a(n) for n = 0..200
Index entries for linear recurrences with constant coefficients, signature (3,3,-9).

Cf. A138587, A107767 (partial sums).

CoefficientList[Series[x^2/((3*x - 1)*(3*x^2 - 1)), {x, 0, 50}], x] (* G. C. Greubel, Jul 03 2016 *)
LinearRecurrence[{3,3,-9},{0,0,1},30] (* Harvey P. Dale, Nov 05 2017 *)

Vec(x^2/((3*x-1)*(3*x^2-1))+O(x^99)) \\ Charles R Greathouse IV, Jun 29 2011

a(2*n+1) = 3*a(2*n).
a(2*n) = A122006(2*n)/2.
a(n) = 3*a(n-1) + 3*a(n-4) - 9*a(n-3).
a(n+1) - a(n) = A122006(n).
a(n) = (3^n - A108411(n+1))/6.
G.f.: x^2/((3*x-1)*(3*x^2-1)).
From Colin Barker, Sep 23 2016: (Start)
a(n) = 3^(n-1)/2-3^(n/2-1)/2 for n even.
a(n) = 3^(n-1)/2-3^(n/2-1/2)/2 for n odd.
(End)

Formulae corrected by Johannes W. Meijer, Jun 28 2011

A171496 a(n) = 6a(n-1) - 8a(n-2) for n > 1; a(0) = 6, a(1) = 28.

Original entry on oeis.org

6, 28, 120, 496, 2016, 8128, 32640, 130816, 523776, 2096128, 8386560, 33550336, 134209536, 536854528, 2147450880, 8589869056, 34359607296, 137438691328, 549755289600, 2199022206976, 8796090925056, 35184367894528

Offset: 0

Klaus Brockhaus, Dec 10 2009

Binomial transform of A171495; second binomial transform of A171494; third binomial transform of A010726.

Vincenzo Librandi, Table of n, a(n) for n = 0..500
Index entries for linear recurrences with constant coefficients, signature (6,-8).

Equals 2*A171499.

Cf. A171476, A006516, A010036, A171494, A171495, A010726, A171472, A171473.

[8*4^n-2*2^n: n in [0..30]]; // Vincenzo Librandi, Jul 18 2011

LinearRecurrence[{6,-8},{6,28},30] (* Harvey P. Dale, Dec 21 2014 *)

{m=22; v=concat([6, 28], vector(m-2)); for(n=3, m, v[n]=6*v[n-1]-8*v[n-2]); v}

a(n) = 8*4^n - 2*2^n.
G.f.: 2*(3-4*x)/((1-2*x)*(1-4*x)).
a(n) = A171476(n+1) = A006516(n+2).
a(n+1) - a(n) = A010036(n+2).
a(n) = 4*a(n-1)+2^(n+1) (with a(0)=6). - Vincenzo Librandi, Dec 04 2010
E.g.f.: 2*exp(2*x)*(2*exp(2*x) - 1)*(2*exp(2*x) + 1). - Stefano Spezia, Dec 10 2021

A266214 Numbers n that are not coprime to the numerator of zeta(2n)/(Pi^(2n)).

Original entry on oeis.org

14, 22, 26, 28, 30, 38, 42, 44, 46, 50, 52, 54, 56, 58, 60, 62, 70, 74, 76, 78, 82, 84, 86, 88, 90, 92, 94, 98, 100, 102, 104, 106, 108, 110, 112, 114, 116, 118, 120, 122, 124, 126, 134, 138, 140, 142, 146, 148, 150, 152, 154, 156, 158, 162, 164, 166, 168, 170

Offset: 1

Chris Boyd, Robert Israel, Dec 24 2015

Equivalently, n not coprime to the numerator of 2^(2n-1)*Bernoulli(2n)/(2n)! (see Lekraj Beedassy comment in A046988).
Conjecture 1: for n>=1, a(n) is identical to 2*A072823(n+1).
Conjecture 2: The corresponding GCDs are powers of 2.
Verified for n <= 10000, e.g.,
GCD = 2 for 14, 22, 26, 28, 30, 38, 42, 44, 46, 50, 52, 54, 56, 58, ...
GCD = 4 for 60, 92, 108, 116, 120, 124, 156, 172, 180, 184, 188, ...
GCD = 8 for 248, 376, 440, 472, 488, 496, 504, 632, 696, 728, 744, ...
GCD = 16 for 1008, 1520, 1776, 1904, 1968, 2000, 2016, 2032, 2544, ...
GCD = 32 for 4064, 6112, 7136, 7648, 7904, 8032, 8096, 8128, 8160
Taking GCDs vertically, column 1 = "14, 60, 248, 1008, 4064, ..." appears to be essentially the same as A171499 and A131262; (ii) column 2 = "22, 92, 376, 1520, 6112, ..." appears to be essentially the same as A010036.
From Chris Boyd, Jan 25 2016: (Start)
Determining whether n is a term of this sequence can be approached by considering odd and even factors separately, and exploiting the fact that numerator(zeta(2n)/(Pi^(2n))) = numerator(2^(2n-2)*N_2n/(D_2n*(2n)!)), where N_2n and D_2n are odd coprime integers such that Bernoulli(2n) = N_2n/2D_2n.
Case 1: odd factors. n is a term if it has an odd prime factor p (necessarily irregular) that divides N_2n at a higher multiplicity than it divides (2n)!. No such factor p of N_2n up to 2n = 10000 is of sufficient multiplicity, and the apparent scarcity of squared and higher power factors of N_2n values (see A090997) suggests that no such p is likely to exist.
Case 2: even factors. An even n is a term if 2 divides 2^(2n-2) at a higher multiplicity than it divides (2n)!. The multiplicity of 2 in 2^(2n-2) is 2n-2, and in (2n)! is 2n minus the number of 1's in the binary expansion of 2n (see A005187). Qualifying n values are therefore those where the number of 1's in the binary expansion of 2n is greater than 2. Except for its first term, A072823 comprises integers with three or more 1-bits in their binary expansion. It follows that for m > 1, 2*A072823(m) values belong to this sequence.
In summary, this sequence is essentially a supersequence of 2*A072823. Conjectures 1 and 2 are true if there are no irregular odd primes p that divide n and the numerator of Bernoulli(2n)/(2n)!. (End)

Chris Boyd, Table of n, a(n) for n = 1..10000
C. Boyd in reply to R. Israel, A046988 query, SeqFan list, Dec 22 2015.

Cf. A010036, A072823, A131262, A171499.

select(n -> igcd(n,numer(2^(2*n-1)*bernoulli(2*n)/(2*n)!)) > 1), [$1..1000]);

Select[Range@ 170, ! CoprimeQ[#, Numerator[Zeta[2 #]/Pi^(2 #)]] &] (* Michael De Vlieger, Dec 24 2015 *)

test(n) = if(gcd(numerator(2^(2*n-1)*bernfrac(2*n)/(2*n)!),n)!=1,1,0)
for(i=1,200,if(test(i),print1(i,", ")))

A087079 Number of lunar partitions of n: number of ways of writing n as a lunar sum of distinct terms, ignoring order.

Original entry on oeis.org

1, 1, 2, 4, 8, 16, 32, 64, 128, 256, 1, 5, 22, 92, 376, 1520, 6112, 24512, 98176, 392960, 2, 22, 200, 1696, 13952, 113152, 911360, 7315456, 58621952, 469368832, 4, 92, 1696, 28928, 477184, 7749632, 124911616, 2005925888, 32153534464, 514926313472, 8

Offset: 0

Marc LeBrun, Oct 09 2003

nonn

Without the condition that the numbers are distinct the answers are infinite because 1+1+1+...+1 = 1 in lunar arithmetic - see A087061.

			a(5) = 16: we can write 5 = 5 + any subset of {4, 3, 2, 1} (16 ways).
a(12) = 22: we can write 12 = 12 + any subset of {11, 10, 2, 1} (16 ways), 12 = 2 + 11 + 10 = 2 + 11 = 2 + 10 and those three with 1 added (6 ways).

D Applegate and N. J. A. Sloane, Table of n, a(n) for n = 0..2000
D. Applegate, C program for lunar arithmetic and number theory [Note: we have now changed the name from "dismal arithmetic" to "lunar arithmetic" - the old name was too depressing]
D. Applegate, M. LeBrun and N. J. A. Sloane, Dismal Arithmetic, arXiv:1107.1130 [math.NT], 2011. [Note: we have now changed the name from "dismal arithmetic" to "lunar arithmetic" - the old name was too depressing]
D. Applegate, M. LeBrun and N. J. A. Sloane, Dismal Arithmetic, J. Int. Seq. 14 (2011) # 11.9.8.
Index entries for sequences related to dismal (or lunar) arithmetic

Cf. A010036.

The subsequence a(n) where n = 111..11 is A003465. - N. J. A. Sloane, May 21 2011

A087079(n) = { my(v, r = 0, i, j, b); v = select(x -> x != 0, digits(n)); for (i = 0, 2^#v - 1, b = Vecrev(binary(i)); b = vector(#v, i, if (i <= #b, b[i], 0)); r += (-1)^vecsum(b) * 2^prod(j = 1, #v, if (b[j] == 1, v[j], v[j] + 1)); ); r/2;} /* Jerome Raulin, Feb 15 2017 */

For 1 <= a < 10 and 0 <= b < 10, a(10a+b) = 2^(ab+a+b-1)+(2^a-1)(2^b-1)2^(ab-1). - David Wasserman, Apr 14 2005

More terms from David Wasserman, Apr 14 2005

A093357 Number of occurrences of pattern 2-1 after n iterations of morphism A007413.

Original entry on oeis.org

0, 4, 20, 88, 368, 1504, 6080, 24448, 98048, 392704, 1571840, 6289408, 25161728, 100655104, 402636800, 1610579968, 6442385408, 25769672704, 103078952960, 412316336128, 1649266393088, 6597067669504, 26388274872320

Offset: 1

Ralf Stephan, Apr 27 2004

S. Kitaev and T. Mansour, Counting the occurrences of generalized patterns....
Index entries for linear recurrences with constant coefficients, signature (6, -8).

Cf. A007413, A010036.

Join[{0},Table[(3*4^(n-1)-2^n)/2,{n,2,30}]] (* or *) Join[{0}, LinearRecurrence[{6,-8},{4,20},30]] (* Harvey P. Dale, Apr 04 2012 *)

PARI
```
a(n)=if(n==1,0,(3*4^(n-1)-2^n)/2)
```

a(1) = 0, a(n) = (3*4^(n-1) - 2^n)/2.
G.f.: 4*x*(1-x)/((1-2*x)*(1-4*x)).
a(1)=0, a(2)=4, a(3)=20, a(n)=6*a(n-1)-8*a(n-2). - Harvey P. Dale, Apr 04 2012
a(n) = 4*A010036(n-2). - R. J. Mathar, Apr 07 2022

A109765 Expansion of x/((4x-1)(2x-1)(x+1)).

Original entry on oeis.org

0, 1, 5, 23, 97, 399, 1617, 6511, 26129, 104687, 419089, 1677039, 6709521, 26840815, 107368721, 429485807, 1717965073, 6871903983, 27487703313, 109950988015, 439804301585, 1759217905391, 7036873019665, 28147494874863

Offset: 0

Creighton Dement, Aug 13 2005

In reference to the program code given, 1baseksumseq[C*D] = A001045 (Jacobsthal sequence, disregard signs).

Floretion Algebra Multiplication Program, FAMP Code: 1basejsumseq[C*D] with C = - 'j + 'k - j' + k' - 'ii' - 'ij' - 'ik' - 'ji' - 'ki' and D = + .5'i + .5'k - .5j' - .5k' + .5'ii' + .5'jj' + .5'jk' + .5'ki'; sumtype: sum[Y[15]] = sum[Y[ * ]], disregard signs

Index entries for linear recurrences with constant coefficients, signature (5,-2,-8).

Cf. A001045, A006516, A010036, A006095.

CoefficientList[Series[x/((4x-1)(2x-1)(x+1)),{x,0,30}],x] (* or *)
LinearRecurrence[{5,-2,-8},{0,1,5},30] (* Harvey P. Dale, Jan 02 2013 *)

a(n) = 5*a(n-1) - 2*a(n-2) - 8*a(n-3), n >= 3.
a(n) = (1/15)*(6*4^n-5*2^n-(-1)^n).
a(n+1) + a(n) = A006516(n+1).
a(n+2) - a(n) = A010036(n+1).

A121544 Sum of all proper base 4 numbers with n digits (those not beginning with 0).

Original entry on oeis.org

6, 114, 1896, 30624, 491136, 7862784, 125822976, 2013241344, 32212156416, 515395682304, 8246335635456, 131941389041664, 2111062300164096, 33776997104615424, 540431954881806336, 8646911282940739584, 138350580546379186176, 2213609288819376390144

Offset: 1

Jonathan Vos Post, Sep 08 2006

Sum of the first 3 * 4^(n-1) integers starting with 4^(n-1).
Sum of the integers from 4^(n-1) to 4^n -1.
First differences of A026337.

			a(1) = 6 = 1 + 2 + 3.
a(2) = 114 = 10_4 + 11_4 + 12_4 + 13_4 + 20_4 + 21_4 + 22_4 + 23_4 + 30_4 + 31_4 + 32_4 + 33_4 = (4+5+6+7+8+9+10+11+12+13+14+15)_10.

G. C. Greubel, Table of n, a(n) for n = 1..820
Index entries for linear recurrences with constant coefficients, signature (20,-64).

Cf. A007090, A010036, A026121, A026337, A101291.

[3*Binomial(5*4^(n-1), 2)/5: n in [1..20]]; // G. C. Greubel, Nov 07 2024

Table[3*4^(n-1)*(5*4^(n-1) - 1)/2, {n,20}] (* James C. McMahon, Oct 19 2024 *)

def A121544(n): return 3*binomial(5*4^(n-1), 2)//5
[A121544(n) for n in range(1,21)] # G. C. Greubel, Nov 07 2024

a(n) = 3 * 4^(n-1) * (4^(n-1) + 4^n - 1)/2.
G.f.: 6*x*(1-x) / ((1-4*x)*(1-16*x)). - Colin Barker, Apr 30 2013
From G. C. Greubel, Nov 07 2024: (Start)
a(n) = (3/5)*binomial(5*4^(n-1), 2).
E.g.f.: (3/32)*(-1 - 4*exp(4*x) + 5*exp(16*x)). (End)

More terms from Colin Barker, Apr 30 2013

Edited by Michel Marcus, Apr 15 2024

A226508 a(n) = Sum_{i=3^n..3^(n+1)-1} i.

Original entry on oeis.org

3, 33, 315, 2889, 26163, 235953, 2125035, 19129689, 172180323, 1549662273, 13947078555, 125524061289, 1129717614483, 10167461718993, 91507165036875, 823564514029689, 7412080712360643, 66708726669526113, 600378540800575995, 5403406869529706889

Offset: 0

Michel Marcus, Jun 10 2013

Partial sums give 3, 36, 351, 3240, 29403,...: A026121.

a(n) is the sum of all integers having n+1 digits in their ternary expansion (without leading zeros). - Jonathan Vos Post, Sep 07 2006

			a(0) = 1+2 = 3,
a(1) = 3+4+5+6+7+8 = 33.

Index entries for linear recurrences with constant coefficients, signature (12,-27).

Cf. A010035, A010036 (base 2), A026121, A101291 (base 10).

Cf. A007089 (numbers in base 3).

Table[3^(n - 1) (4 3^(n + 1) - 3), {n, 0, 20}] (* Bruno Berselli, Jun 11 2013 *)
LinearRecurrence[{12,-27},{3,33},30] (* Harvey P. Dale, Jun 19 2013 *)

a(n) = sum(i=3^n, 3^(n+1)-1, i) \\ Michel Marcus, Jun 11 2013

G.f.: 3*(1-x)/(1-12*x+27*x^2). [Bruno Berselli, Jun 11 2013]
a(n) = 3^(n-1)*(4*3^(n+1)-3). [Bruno Berselli, Jun 11 2013]
a(0)=3, a(1)=33, a(n)=12*a(n-1)-27*a(n-2). - Harvey P. Dale, Jun 19 2013

Previous Showing 11-20 of 26 results. Next

cp's OEIS Frontend

A165665 a(n) = (3*2^n - 2) * 2^n.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Magma

Mathematica

PARI

Formula

A125185 Triangle read by rows: T(n,k) is the coefficient of t^k in the polynomial S(n,t)=[(1+t)(2+t)^n+(1-t)t^n]/2 (0<=k<=n).

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Maple

Formula

A167993 Expansion of x^2/((3*x-1)*(3*x^2-1)).

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Mathematica

PARI

Formula

Extensions

A171496 a(n) = 6*a(n-1) - 8*a(n-2) for n > 1; a(0) = 6, a(1) = 28.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Magma

Mathematica

PARI

Formula

A266214 Numbers n that are not coprime to the numerator of zeta(2*n)/(Pi^(2*n)).

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Maple

Mathematica

PARI

A087079 Number of lunar partitions of n: number of ways of writing n as a lunar sum of distinct terms, ignoring order.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

PARI

Formula

Extensions

A093357 Number of occurrences of pattern 2-1 after n iterations of morphism A007413.

Views

Author

Keywords

Links

Crossrefs

Programs

Mathematica

PARI

A165665 a(n) = (32^n - 2) 2^n.

A167993 Expansion of x^2/((3x-1)(3*x^2-1)).

A171496 a(n) = 6a(n-1) - 8a(n-2) for n > 1; a(0) = 6, a(1) = 28.

A266214 Numbers n that are not coprime to the numerator of zeta(2n)/(Pi^(2n)).

A109765 Expansion of x/((4x-1)(2x-1)(x+1)).