A005704 -id:A005704 - cp's OEIS frontend

A072720 Number of partitions of n into parts which are each powers of a single number (which may vary between partitions).

1, 1, 2, 3, 5, 6, 10, 11, 15, 17, 23, 24, 34, 35, 43, 47, 57, 58, 73, 74, 91, 96, 112, 113, 139, 141, 163, 168, 197, 198, 235, 236, 272, 279, 317, 321, 378, 379, 427, 436, 501, 502, 575, 576, 653, 666, 742, 743, 851, 853, 952, 963, 1080, 1081, 1211, 1216, 1361

Offset: 0

Henry Bottomley, Jul 05 2002

nonn

First differs from A322912 at a(12) = 34, A322912(12) = 33.

			a(6)=10 since 6 can be written as 6 (powers of 6), 5+1 (5), 4+1+1 (4 or 2), 3+3 (3), 3+1+1+1 (3), 4+2 (2), 2+2+2 (2), 2+2+1+1 (2), 2+1+1+1+1 (2) and 1+1+1+1+1+1 (powers of anything).
From _Gus Wiseman_, Jan 01 2019: (Start)
The a(1) = 1 through a(8) = 15 integer partitions:
  (1)  (2)   (3)    (4)     (5)      (6)       (7)        (8)
       (11)  (21)   (22)    (41)     (33)      (61)       (44)
             (111)  (31)    (221)    (42)      (331)      (71)
                    (211)   (311)    (51)      (421)      (422)
                    (1111)  (2111)   (222)     (511)      (611)
                            (11111)  (411)     (2221)     (2222)
                                     (2211)    (4111)     (3311)
                                     (3111)    (22111)    (4211)
                                     (21111)   (31111)    (5111)
                                     (111111)  (211111)   (22211)
                                               (1111111)  (41111)
                                                          (221111)
                                                          (311111)
                                                          (2111111)
                                                          (11111111)
(End)

Cf. A000123, A005704, A005705, A005706, A072721.

Cf. A018819, A023893, A052410, A102430, A322900, A322901, A322902, A322903, A322912.

radbase[n_]:=n^(1/GCD@@FactorInteger[n][[All,2]]);
Table[Length[Select[IntegerPartitions[n],SameQ@@radbase/@DeleteCases[#,1]&]],{n,30}] (* Gus Wiseman, Jan 01 2019 *)

a(n) = a(n-1) + A072721(n). a(p) = a(p-1)+1 for p prime.

A292477 Square array A(n,k), n >= 0, k >= 2, read by antidiagonals: A(n,k) = [x^(k*n)] Product_{j>=0} 1/(1 - x^(k^j)).

Original entry on oeis.org

1, 1, 2, 1, 2, 4, 1, 2, 3, 6, 1, 2, 3, 5, 10, 1, 2, 3, 4, 7, 14, 1, 2, 3, 4, 6, 9, 20, 1, 2, 3, 4, 5, 8, 12, 26, 1, 2, 3, 4, 5, 7, 10, 15, 36, 1, 2, 3, 4, 5, 6, 9, 12, 18, 46, 1, 2, 3, 4, 5, 6, 8, 11, 15, 23, 60, 1, 2, 3, 4, 5, 6, 7, 10, 13, 18, 28, 74, 1, 2, 3, 4, 5, 6, 7, 9, 12, 15, 21, 33, 94

Offset: 0

Ilya Gutkovskiy, Sep 17 2017

A(n,k) is the number of partitions of k*n into powers of k.

			Square array begins:
   1,  1,  1,  1,  1,  1, ...
   2,  2,  2,  2,  2,  2, ...
   4,  3,  3,  3,  3,  3, ...
   6,  5,  4,  4,  4,  4, ...
  10,  7,  6,  5,  5,  5, ...
  14,  9,  8,  7,  6,  6, ...

Seiichi Manyama, Antidiagonals n = 0..139, flattened
George E. Andrews, Aviezri S. Fraenkel, James A. Sellers, Characterizing the number of m-ary partitions modulo m, Amer. Math. Monthly 122:9 (2015), 880-885.
Index entries for related partition-counting sequences

Columns k=2..5 give A000123, A005704, A005705, A005706.
Mirror of A089688 (excluding the first row).
Cf. A145515, A294316.

Table[Function[k, SeriesCoefficient[Product[1/(1 - x^k^i), {i, 0, n}], {x, 0, k n}]][j - n + 2], {j, 0, 12}, {n, 0, j}] // Flatten

A309677 G.f. A(x) satisfies: A(x) = A(x^3) / (1 - x)^2.

Original entry on oeis.org

1, 2, 3, 6, 9, 12, 18, 24, 30, 42, 54, 66, 87, 108, 129, 162, 195, 228, 279, 330, 381, 456, 531, 606, 711, 816, 921, 1068, 1215, 1362, 1563, 1764, 1965, 2232, 2499, 2766, 3120, 3474, 3828, 4290, 4752, 5214, 5805, 6396, 6987, 7740, 8493, 9246, 10194, 11142

Offset: 0

Ilya Gutkovskiy, Aug 12 2019

nonn

Self-convolution of A062051.

Alois P. Heinz, Table of n, a(n) for n = 0..10000

Cf. A005704, A062051, A120880, A171238, A309678, A309679.

b:= proc(n, i) option remember; `if`(n=0, 1, `if`(i<0, 0,
      b(n, i-1)+(p-> `if`(p>n, 0, b(n-p, i)))(3^i)))
    end:
a:= n-> add(b(j, ilog[3](j))*b(n-j, ilog[3](n-j)), j=0..n):
seq(a(n), n=0..52);  # Alois P. Heinz, Aug 12 2019

nmax = 52; A[] = 1; Do[A[x] = A[x^3]/(1 - x)^2 + O[x]^(nmax + 1) // Normal, nmax + 1]; CoefficientList[A[x], x]
nmax = 52; CoefficientList[Series[Product[1/(1 - x^(3^k))^2, {k, 0, Floor[Log[3, nmax]] + 1}], {x, 0, nmax}], x]

G.f.: Product_{k>=0} 1/(1 - x^(3^k))^2.

A371709 Expansion of g.f. A(x) satisfying A( xA(x)^2 + xA(x)^3 ) = A(x)^3.

Original entry on oeis.org

1, 1, 1, 2, 6, 16, 39, 99, 271, 764, 2157, 6128, 17658, 51534, 151635, 448962, 1337493, 4008040, 12072594, 36524898, 110943633, 338218626, 1034509917, 3173811240, 9763898994, 30113782641, 93094164244, 288415278638, 895332445053, 2784580242557, 8675408291598, 27072326322939

Offset: 1

Paul D. Hanna, May 02 2024

nonn

Compare to the following identities of the Catalan function C(x) = x + C(x)^2 (A000108):
(1) C(x)^2 = C( x*C(x)*(1 + C(x)) ),
(2) C(x)^4 = C( x*C(x)^3*(1 + C(x))*(1 + C(x)^2) ),
(3) C(x)^8 = C( x*C(x)^7*(1 + C(x))*(1 + C(x)^2)*(1 + C(x)^4) ),
(4) C(x)^(2^n) = C( x*C(x)^(2^n-1)*Product_{k=0..n-1} (1 + C(x)^(2^k)) ) for n > 0.
a(3^n) == 1 (mod 3) for n >= 0.
a(2*3^n) == 1 (mod 3) for n >= 0.
a(n) == 2 (mod 3) iff n is the sum of 2 distinct powers of 3 (A038464).

			G.f. A(x) = x + x^2 + x^3 + 2*x^4 + 6*x^5 + 16*x^6 + 39*x^7 + 99*x^8 + 271*x^9 + 764*x^10 + 2157*x^11 + 6128*x^12 + 17658*x^13 + 51534*x^14 + 151635*x^15 + 448962*x^16 + ...
where A( x*A(x)^2*(1 + A(x)) ) = A(x)^3.
RELATED SERIES.
A(x)^2 = x^2 + 2*x^3 + 3*x^4 + 6*x^5 + 17*x^6 + 48*x^7 + 126*x^8 + 332*x^9 + 918*x^10 + 2616*x^11 + 7504*x^12 + ...
A(x)^3 = x^3 + 3*x^4 + 6*x^5 + 13*x^6 + 36*x^7 + 105*x^8 + 292*x^9 + 801*x^10 + 2256*x^11 + 6515*x^12 + 18981*x^13 + ...
A(x)^2 + A(x)^3 = x^2 + 3*x^3 + 6*x^4 + 12*x^5 + 30*x^6 + 84*x^7 + 231*x^8 + 624*x^9 + 1719*x^10 + 4872*x^11 + 14019*x^12 + 40599*x^13 + ...
Let B(x) be the series reversion of g.f. A(x), B(A(x)) = x, then
B(x) * (1+x)/(1+x^3) = x - 2*x^4 + 3*x^7 - 5*x^10 + 7*x^13 - 9*x^16 + 12*x^19 - 15*x^22 + 18*x^25 - 23*x^28 + ... + (-1)^n*A005704(n)*x^(3*n+1) + ...
where A005704 is the number of partitions of 3*n into powers of 3.
We can show that g.f. A(x) = A( x*A(x)^2*(1 + A(x)) )^(1/3) satisfies
(4) A(x) = x * Product_{n>=0} (1 + A(x)^(3^n))
by substituting x*A(x)^2*(1 + A(x)) for x in (4) to obtain
A(x)^3 = x * A(x)^2*(1 + A(x)) * Product_{n>=1} (1 + A(x)^(3^n))
which is equivalent to formula (4).
SPECIFIC VALUES.
A(3/10) = 0.526165645044542830201162330432965674027415264612114520...
A(1/4) = 0.353259384374080248921564026412797625837830114153200664...
A(1/5) = 0.255218141344695821239609680309162895225297482063273545...
A(t) = 1/2 and A(t*3/8) = 1/8 at t = (1/2)/Product_{n>=0} (1 + 1/2^(3^n)) = 0.295718718466711580562679377308518930409875701753934072...
A(t) = 1/3 and A(t*4/27) = 1/27 at t = (1/3)/Product_{n>=0} (1 + 1/3^(3^n)) = 0.241059181496179959557718992589733756750585121455883861...
A(t) = 1/4 and A(t*5/64) = 1/64 at t = (1/4)/Product_{n>=0} (1 + 1/4^(3^n)) = 0.196922325724019432212969925740117827612003158137366017...

Paul D. Hanna, Table of n, a(n) for n = 1..1000

Cf. A371716, A370440, A370441, A370446, A264228, A198951.

Cf. A005704, A038464.

/* Using series reversion of x/Product_{n>=0} (1 + x^(3^n)) */
{a(n) = my(A); A = serreverse( x/prod(k=0,ceil(log(n)/log(3)), (1 + x^(3^k) +x*O(x^n)) ) ); polcoeff(A,n)}
for(n=1,35, print1(a(n),", "))

/* Using A(x)^3 = A( x*A(x)^2 + x*A(x)^3 ) */
{a(n) = my(A=[1],F); for(i=1,n, A = concat(A,0); F = x*Ser(A);
A[#A] = polcoeff( subst(F,x, x*F^2 + x*F^3 ) - F^3, #A+2) ); A[n]}
for(n=1,35, print1(a(n),", "))

G.f. A(x) = Sum_{n>=1} a(n)*x^n satisfies the following formulas.
(1) A(x)^3 = A( x*A(x)^2*(1 + A(x)) ).
(2) A(x)^9 = A( x*A(x)^8*(1 + A(x))*(1 + A(x)^3) ).
(3) A(x)^27 = A( x*A(x)^26*(1 + A(x))*(1 + A(x)^3)*(1 + A(x)^9) ).
(4) A(x) = x * Product_{n>=0} (1 + A(x)^(3^n)).
(5) A(x) = Series_Reversion( x / Product_{n>=0} (1 + x^(3^n)) ).
a(n) ~ c * d^n / n^(3/2), where d = 3.2753449994351908157330968510747739... and c = 0.1559869008021616116037651076359... - Vaclav Kotesovec, May 03 2024
The radius of convergence r of g.f. A(x) and A(r) satisfy 1 = Sum_{n>=0} 3^n * A(r)^(3^n) / (1 + A(r)^(3^n)) and r = A(r) / Product_{n>=0} (1 + A(r)^(3^n)), where r = 0.30531134893345362211... = 1/d (d is given above) and A(r) = 0.600582105427215700175254768411726892599... - Paul D. Hanna, May 03 2024

A089649 a(1)=1, a(2)=2, a(n) = a(n-1) + a(floor((n+1)/3)).

Original entry on oeis.org

1, 2, 3, 4, 6, 8, 10, 13, 16, 19, 23, 27, 31, 37, 43, 49, 57, 65, 73, 83, 93, 103, 116, 129, 142, 158, 174, 190, 209, 228, 247, 270, 293, 316, 343, 370, 397, 428, 459, 490, 527, 564, 601, 644, 687, 730, 779, 828, 877, 934, 991, 1048, 1113, 1178, 1243, 1316, 1389

Offset: 1

Philippe Deléham, Jan 02 2004

nonn

Partial sums of the sequence: a(1)=1, a(1), a(1), a(1), a(2), a(2), a(2), a(3), a(3), ... each term repeated 3 times.

Ivan Neretin, Table of n, a(n) for n = 1..10000

Cf. A005704, A089650.

Fold[Append[#1, #1[[-1]] + #1[[Quotient[#2 + 1, 3]]]] &, {1, 2}, Range[3, 57]] (* Ivan Neretin, Nov 25 2016 *)

PARI
```
a(n)=if(n<2,1,a(n-1)+a(floor((n+1)/3)))
```

A183038 G.f.: exp( Sum_{n>=1} A051064(n)3^A051064(n)x^n/n ) where A051064(n) equals the 3-adic valuation of 3n.

Original entry on oeis.org

1, 3, 6, 15, 30, 51, 93, 156, 240, 387, 597, 870, 1311, 1920, 2697, 3873, 5448, 7422, 10278, 14016, 18636, 25098, 33402, 43548, 57333, 74757, 95820, 123780, 158637, 200391, 254778, 321798, 401451, 503490, 627915, 774726, 960156, 1184205, 1446873

Offset: 0

Paul D. Hanna, Dec 19 2010

nonn

Compare to B(x), the g.f. of the number of partitions of 3n into powers of 3 (A005704):

B(x) = exp( Sum_{n>=1} 3^A051064(n)*x^n/n ) = (1-x)^(-1)*Product_{n>=0} 1/(1 - x^(3^n)).

			G.f.: A(x) =  1 + 3*x + 6*x^2 + 15*x^3 + 30*x^4 + 51*x^5 + 93*x^6 +...
log(A(x)) = 3*x + 3*x^2/2 + 18*x^3/3 + 3*x^4/4 + 3*x^5/5 + 18*x^6/6 + 3*x^7/7 + 3*x^8/8 + 81*x^9/9 + 3*x^10/10 + 3*x^11/11 + 18*x^12/12 +...
G.f. satisfies A(x) = A(x^3)*G(x) where G(x) is the g.f. of A161809:
G(x) = 1 + 3*x + 6*x^2 + 12*x^3 + 21*x^4 + 33*x^5 + 51*x^6 +...
TRISECTIONS of g.f. begin:
T_0(x) = 1 + 15*x + 93*x^2 + 387*x^3 + 1311*x^4 + 3873*x^5 +...
T_1(x) = 3 + 30*x + 156*x^2 + 597*x^3 + 1920*x^4 + 5448*x^5 +...
T_2(x) = 6 + 51*x + 240*x^2 + 870*x^3 + 2697*x^4 + 7422*x^5 +...
where the ratios involve Fibonacci numbers:
T_1(x)/T_0(x) = 3*(1 - 5*x + 34*x^2 - 233*x^3 +...+ (-1)^n*Fibonacci(4n+1)*x^n +...);
T_2(x)/T_0(x) = 3*(2 - 13*x + 89*x^2 - 610*x^3 +...+ (-1)^n*Fibonacci(4n+3)*x^n +...).

Cf. A161809, A183039, A051064, A005704, A183036.

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

G.f. satisfies: A(x) = (1-x^3)/(1-x)^3 * A(x^3)^2/A(x^9).
G.f. satisfies: A(x) = A(x^3)*G(x) where G(x) = G(x^3)*(1+x+x^2)/(1-x)^2 is the g.f. of A161809.
Define TRISECTIONS: A(x) = T_0(x^3) + x*T_1(x^3) + x^2*T_2(x^3), then:
T_1(x)/T_0(x) = 3*(1 + 2*x)/(1 + 7*x + x^2) and
T_2(x)/T_0(x) = 3*(2 + x)/(1 + 7*x + x^2).

A268128 a(n) = (A000123(n) - A001316)/2.

Original entry on oeis.org

0, 0, 1, 1, 4, 5, 8, 9, 17, 21, 28, 33, 45, 53, 66, 75, 100, 117, 140, 161, 193, 221, 258, 291, 344, 389, 446, 499, 573, 639, 722, 797, 913, 1013, 1132, 1249, 1393, 1533, 1698, 1859, 2060, 2253, 2478, 2699, 2965, 3223, 3522, 3813, 4173, 4517, 4910, 5299, 5753

Offset: 0

Tom Edgar, Jan 26 2016

nonn

G. E. Andrews, A. S. Fraenkel, and J. A. Sellers, Characterizing the number of m-ary partitions modulo m, The American Mathematical Monthly, Vol. 122, No. 9 (November 2015), pp. 880-885.
G. E. Andrews, A. S. Fraenkel, and J. A. Sellers, Characterizing the number of m-ary partitions modulo m.
Tom Edgar, The distribution of the number of parts of m-ary partitions modulo m., arXiv:1603.00085 [math.CO], 2016.

Cf. A005704, A006047, A268127.

b[0] = 1; b[n_] := b[n] = b[Floor[n/2]] + b[n - 1];
c[n_] := Sum[Mod[Binomial[n, k], 2], {k, 0, n}];
a[n_] := (b[n] - c[n])/2;
Table[a[n], {n, 0, 52}] (* Jean-François Alcover, Dec 12 2018 *)

def b(n):
    A=[1]
    for i in [1..n]:
        A.append(A[i-1] + A[floor(i/2)])
    return A[n]
[(b(n)-prod(x+1 for x in n.digits(2)))/2 for n in [0..60]]

Let b(0) = 1 and b(n) = b(n-1) + b(floor(n/2)) and let c(n) = Product_{i=0..k}(n_i+1) where n = Sum_{i=0..k}n_i*2^i is the binary representation of n. Then a(n) = (1/2)*(b(n) - c(n)).

A294298 Sum of products of terms in all partitions of 3*n into powers of 3.

Original entry on oeis.org

1, 4, 13, 49, 157, 481, 1534, 4693, 14170, 43357, 130918, 393601, 1188454, 3573013, 10726690, 32248957, 96815758, 290516161, 872169223, 2617128409, 7852005967, 23561605318, 70690403371, 212076797530, 636280680100, 1908892327810, 5726727270940, 17180634420931

Offset: 0

Seiichi Manyama, Oct 27 2017

nonn

			n | partitions of 3*n into powers of 3                          | a(n)
----------------------------------------------------------------------------------
1 | 3  , 1+1+1                                                  | 3+1        =  4.
2 | 3+3, 3+1+1+1, 1+1+1+1+1+1                                   | 9+3+1      = 13.
3 | 9  , 3+3+3  , 3+3+1+1+1  , 3+1+1+1+1+1+1, 1+1+1+1+1+1+1+1+1 | 9+27+9+3+1 = 49.

Seiichi Manyama, Table of n, a(n) for n = 0..1000

Cf. A005704, A062051, A289842.

b:= proc(n, i, p) option remember; `if`(n=0, p,
     `if`(i<1, 0, add(b(n-j*i, i/3, p*i^j), j=0..n/i)))
    end:
a:= n-> (t-> b(t, 3^ilog[3](t), 1))(3*n):
seq(a(n), n=0..33);  # Alois P. Heinz, Oct 27 2017

b[n_, i_, p_] := b[n, i, p] = If[n == 0, p, If[i < 1, 0, Sum[b[n - j i, i/3, p i^j], {j, 0, n/i}]]];
a[n_] := b[3n, 3^Floor@Log[3, 3n], 1];
a /@ Range[0, 33] (* Jean-François Alcover, Nov 23 2020, after Alois P. Heinz *)

a(n) = [x^(3*n)] Product_{k>=0} 1/(1 - 3^k*x^(3^k)). - Ilya Gutkovskiy, Sep 10 2018

a(n) ~ c * 3^n, where c = 2.2530906593645919365992433370928351696108819534655299832797806149219665... - Vaclav Kotesovec, Jun 18 2019

A089688 Table T(n,k), n>=0 and k>=1, read by antidiagonals; the k-th row is defined by : partitions of k*n into powers of k (with T(0,k) = 1).

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 4, 2, 1, 1, 6, 3, 2, 1, 1, 10, 5, 3, 2, 1, 1, 14, 7, 4, 3, 2, 1, 1, 20, 9, 6, 4, 3, 2, 1, 1, 26, 12, 8, 5, 4, 3, 2, 1, 1, 36, 15, 10, 7, 5, 4, 3, 2, 1, 1, 46, 18, 12, 9, 6, 5, 4, 3, 2, 1, 1, 60, 23, 15, 11, 8, 6, 5, 4, 3, 2, 1

Offset: 0

Philippe Deléham, Jan 05 2004

			Row k = 1 : 1, 1, 1, 1,  1,  1,  1,  1,  1,  1,  1,  1, ... (see A000012).
Row k = 2 : 1, 2, 4, 6, 10, 14, 20, 26, 36, 46, 60, 74, ... (see A000123).
Row k = 3 : 1, 2, 3, 5,  7,  9, 12, 15, 18, 23, 28, 33, ... (see A005704).
Row k = 4 : 1, 2, 3, 4,  6,  8, 10, 12, 15, 18, 21, 24, ... (see A005705).
Row k = 5 : 1, 2, 3, 4,  5,  7,  9, 11, 13, 15, 18, 21, ... (see A005706).

Alois P. Heinz, Table of n, a(n) for n = 0..10010

A306727 Square array A(n,k), n >= 0, k >= 0, read by antidiagonals: A(n,k) is the number of partitions of 3*n into powers of 3 less than or equal to 3^k.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 2, 3, 1, 1, 2, 3, 4, 1, 1, 2, 3, 5, 5, 1, 1, 2, 3, 5, 7, 6, 1, 1, 2, 3, 5, 7, 9, 7, 1, 1, 2, 3, 5, 7, 9, 12, 8, 1, 1, 2, 3, 5, 7, 9, 12, 15, 9, 1, 1, 2, 3, 5, 7, 9, 12, 15, 18, 10, 1, 1, 2, 3, 5, 7, 9, 12, 15, 18, 22, 11, 1, 1, 2, 3, 5, 7, 9, 12, 15, 18, 23, 26, 12, 1

Offset: 0

Serguei Zolotov, Mar 06 2019

Column sequences converge to A005704.

			A(3,3) = 5, because there are 5 partitions of 3*3=9 into powers of 3 less than or equal to 3^3=9: [9], [3,3,3], [3,3,1,1,1], [3,1,1,1,1,1,1], [1,1,1,1,1,1,1,1,1].
Square array A(n,k) begins:
  1, 1, 1, 1, 1, 1,  ...
  1, 2, 2, 2, 2, 2,  ...
  1, 3, 3, 3, 3, 3,  ...
  1, 4, 5, 5, 5, 5,  ...
  1, 5, 7, 7, 7, 7,  ...
  1, 6, 9, 9, 9, 9,  ...

Serguei Zolotov, Table of n, a(n) for n = 0..10584

Main diagonal gives A005704.

A181322 gives array for base p=2.

nmax = 12;
f[k_] := f[k] = 1/(1-x) 1/Product[1-x^(3^j), {j, 0, k-1}] + O[x]^(nmax+1) // CoefficientList[#, x]&;
A[n_, k_] := f[k][[n+1]];
Table[A[n-k, k], {n, 0, nmax}, {k, n, 0, -1}] // Flatten (* Jean-François Alcover, Nov 20 2019 *)

def aseq(p, x, k):
    # generic algorithm for any p - power base, p=3 for this sequence
    if x < 0:
        return 0
    if x < p:
        return 1
    # coefficients
    arr = [0]*(x+1)
    arr[0] = 1
    m = p**k
    while m > 0:
        for i in range(m, x+1, m):
            arr[i] += arr[i-m]
        m //= p
    return arr[x]
def A(n, k):
    p = 3
    return aseq(p, p*n, k)
# A(n, k), 5 = A(3, 3) = aseq(3, 3*3, 3)
# Serguei Zolotov, Mar 13 2019

G.f. of column k: 1/(1-x) * 1/Product_{j=0..k-1} (1 - x^(3^j)).

cp's OEIS Frontend

A072720 Number of partitions of n into parts which are each powers of a single number (which may vary between partitions).

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A292477 Square array A(n,k), n >= 0, k >= 2, read by antidiagonals: A(n,k) = [x^(k*n)] Product_{j>=0} 1/(1 - x^(k^j)).

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A309677 G.f. A(x) satisfies: A(x) = A(x^3) / (1 - x)^2.

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Maple

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A371709 Expansion of g.f. A(x) satisfying A( x*A(x)^2 + x*A(x)^3 ) = A(x)^3.

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PARI

PARI

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A089649 a(1)=1, a(2)=2, a(n) = a(n-1) + a(floor((n+1)/3)).

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A183038 G.f.: exp( Sum_{n>=1} A051064(n)*3^A051064(n)*x^n/n ) where A051064(n) equals the 3-adic valuation of 3n.

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A268128 a(n) = (A000123(n) - A001316)/2.

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Sage

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A294298 Sum of products of terms in all partitions of 3*n into powers of 3.

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A371709 Expansion of g.f. A(x) satisfying A( xA(x)^2 + xA(x)^3 ) = A(x)^3.

A183038 G.f.: exp( Sum_{n>=1} A051064(n)3^A051064(n)x^n/n ) where A051064(n) equals the 3-adic valuation of 3n.