A268685 -id:A268685 - cp's OEIS frontend

A319014 a(n) = 123 + 456 + 789 + 101112 + 131415 + 161718 + ... + (up to n).

1, 2, 6, 10, 26, 126, 133, 182, 630, 640, 740, 1950, 1963, 2132, 4680, 4696, 4952, 9576, 9595, 9956, 17556, 17578, 18062, 29700, 29725, 30350, 47250, 47278, 48062, 71610, 71641, 72602, 104346, 104380, 105536, 147186, 147223, 148592, 202020, 202060, 203660

Offset: 1

Wesley Ivan Hurt, Sep 07 2018

In general, for sequences that multiply the first k natural numbers, and then add the product of the next k natural numbers (preserving the order of operations up to n), we have a(n) = Sum_{i=1..floor(n/k)} (k*i)!/(k*i-k)! + Sum_{j=1..k-1} (1-sign((n-j) mod k)) * (Product_{i=1..j} n-i+1). Here, k=3.

			a(1)  = 1;
a(2)  = 1*2 = 2;
a(3)  = 1*2*3 = 6;
a(4)  = 1*2*3 + 4 = 10;
a(5)  = 1*2*3 + 4*5 = 26;
a(6)  = 1*2*3 + 4*5*6 = 126;
a(7)  = 1*2*3 + 4*5*6 + 7 = 133;
a(8)  = 1*2*3 + 4*5*6 + 7*8 = 182;
a(9)  = 1*2*3 + 4*5*6 + 7*8*9 = 630;
a(10) = 1*2*3 + 4*5*6 + 7*8*9 + 10 = 640;
a(11) = 1*2*3 + 4*5*6 + 7*8*9 + 10*11 = 740;
a(12) = 1*2*3 + 4*5*6 + 7*8*9 + 10*11*12 = 1950;
a(13) = 1*2*3 + 4*5*6 + 7*8*9 + 10*11*12 + 13 = 1963;
a(14) = 1*2*3 + 4*5*6 + 7*8*9 + 10*11*12 + 13*14 = 2132;
a(15) = 1*2*3 + 4*5*6 + 7*8*9 + 10*11*12 + 13*14*15 = 4680;
a(16) = 1*2*3 + 4*5*6 + 7*8*9 + 10*11*12 + 13*14*15 + 16 = 4696;
a(17) = 1*2*3 + 4*5*6 + 7*8*9 + 10*11*12 + 13*14*15 + 16*17 = 4952;
a(18) = 1*2*3 + 4*5*6 + 7*8*9 + 10*11*12 + 13*14*15 + 16*17*18 = 9576;
a(19) = 1*2*3 + 4*5*6 + 7*8*9 + 10*11*12 + 13*14*15 + 16*17*18 + 19 = 9595;
etc.

Colin Barker, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (1,0,4,-4,0,-6,6,0,4,-4,0,-1,1).

Cf. A093361, (k=1) A000217, (k=2) A228958, (k=3) this sequence, (k=4) A319205, (k=5) A319206, (k=6) A319207, (k=7) A319208, (k=8) A319209, (k=9) A319211, (k=10) A319212.

Cf. A049347, A061347, A268685 (trisection).

CoefficientList[Series[(1 + x + 4*x^2 + 12*x^4 + 84*x^5 - 3*x^6 - 9*x^7 + 72*x^8 + 2*x^9 - 4*x^10 + 2*x^11)/((1 - x)^5*(1 + x + x^2)^4), {x, 0, 50}], x] (* after Colin Barker *)

Vec(x*(1 + x + 4*x^2 + 12*x^4 + 84*x^5 - 3*x^6 - 9*x^7 + 72*x^8 + 2*x^9 - 4*x^10 + 2*x^11) / ((1 - x)^5*(1 + x + x^2)^4) + O(x^50)) \\ Colin Barker, Sep 08 2018

a(n) = Sum_{i=1..floor(n/3)} (3*i)!/(3*i-3)! + Sum_{j=1..2} (1-sign((n-j) mod 3)) * (Product_{i=1..j} n-i+1).
From Colin Barker, Sep 08 2018: (Start)
G.f.: x*(1 + x + 4*x^2 + 12*x^4 + 84*x^5 - 3*x^6 - 9*x^7 + 72*x^8 + 2*x^9 - 4*x^10 + 2*x^11) / ((1 - x)^5*(1 + x + x^2)^4).
a(n) = a(n-1) + 4*a(n-3) - 4*a(n-4) - 6*a(n-6) + 6*a(n-7) + 4*a(n-9) - 4*a(n-10) - a(n-12) + a(n-13) for n>13.
(End)
a(3*k) = 3*k*(k+1)*(3*k-2)*(3*k+1)/4, a(3*k+1) = a(3*k) + 3*k + 1, a(3*k+2) = a(3*k) + (3*k+2)*(3*k+1). - Giovanni Resta, Sep 08 2018
a(n) = (3*n^4 - 6*n^3 + 9*n^2 + 6*n - 8 - 2*(3*n^3 - 6*n^2 - 6*n + 2)*A061347(n-3) + 6*(n^3 - 6*n^2 + 6*n + 2)*A049347(n-2))/36. - Stefano Spezia, Apr 23 2023

A319867 a(n) = 321 + 654 + 987 + 121110 + ... + (up to the n-th term).

Original entry on oeis.org

3, 6, 6, 12, 36, 126, 135, 198, 630, 642, 762, 1950, 1965, 2160, 4680, 4698, 4986, 9576, 9597, 9996, 17556, 17580, 18108, 29700, 29727, 30402, 47250, 47280, 48120, 71610, 71643, 72666, 104346, 104382, 105606, 147186, 147225, 148668, 202020, 202062, 203742

Offset: 1

Wesley Ivan Hurt, Sep 29 2018

For similar multiply/add sequences in descending blocks of k natural numbers, we have: a(n) = Sum_{j=1..k-1} (floor((n-j)/k)-floor((n-j-1)/k)) * (Product_{i=1..j} n-i-j+k+1) + Sum_{j=1..n} (floor(j/k)-floor((j-1)/k)) * (Product_{i=1..k} j-i+1). Here, k=3.

			a(1) = 3;
a(2) = 3*2 = 6;
a(3) = 3*2*1 = 6;
a(4) = 3*2*1 + 6 = 12;
a(5) = 3*2*1 + 6*5 = 36;
a(6) = 3*2*1 + 6*5*4 = 126;
a(7) = 3*2*1 + 6*5*4 + 9 = 135;
a(8) = 3*2*1 + 6*5*4 + 9*8 = 198;
a(9) = 3*2*1 + 6*5*4 + 9*8*7 = 630;
a(10) = 3*2*1 + 6*5*4 + 9*8*7 + 12 = 642;
a(11) = 3*2*1 + 6*5*4 + 9*8*7 + 12*11 = 762;
a(12) = 3*2*1 + 6*5*4 + 9*8*7 + 12*11*10 = 1950;
a(13) = 3*2*1 + 6*5*4 + 9*8*7 + 12*11*10 + 15 = 1965;
a(14) = 3*2*1 + 6*5*4 + 9*8*7 + 12*11*10 + 15*14 = 2160;
a(15) = 3*2*1 + 6*5*4 + 9*8*7 + 12*11*10 + 15*14*13 = 4680;
a(16) = 3*2*1 + 6*5*4 + 9*8*7 + 12*11*10 + 15*14*13 + 18 = 4698;
a(17) = 3*2*1 + 6*5*4 + 9*8*7 + 12*11*10 + 15*14*13 + 18*17 = 4986;
a(18) = 3*2*1 + 6*5*4 + 9*8*7 + 12*11*10 + 15*14*13 + 18*17*16 = 9576;
a(19) = 3*2*1 + 6*5*4 + 9*8*7 + 12*11*10 + 15*14*13 + 18*17*16 + 21 = 9597;
etc.

Colin Barker, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (1,0,4,-4,0,-6,6,0,4,-4,0,-1,1).

For similar sequences, see: A000217 (k=1), A319866 (k=2), this sequence (k=3), A319868 (k=4), A319869 (k=5), A319870 (k=6), A319871 (k=7), A319872 (k=8), A319873 (k=9), A319874 (k=10).

Cf. A268685 (trisection).

a:=(n,k)->add((floor((n-j)/k)-floor((n-j-1)/k))*(mul(n-i-j+k+1,i=1..j)),j=1..k-1) + add((floor(j/k)-floor((j-1)/k))*(mul(j-i+1,i=1..k)),j=1..n): seq(a(n,3),n=1..45); # Muniru A Asiru, Sep 30 2018

k:=3; a[n_]:=Sum[(Floor[(n-j)/k]-Floor[(n-j-1)/k]) * Product[n-i-j+k+1, {i,1,j }], {j,1,k-1} ] + Sum[(Floor[j/k]-Floor[(j-1)/k]) * Product[j-i+1, {i,1,k}], {j,1,n}]; Array[a, 50] (* Stefano Spezia, Sep 30 2018 *)

Vec(3*x*(1 + x - 2*x^3 + 4*x^4 + 30*x^5 + x^6 - 5*x^7 + 24*x^8) / ((1 - x)^5*(1 + x + x^2)^4) + O(x^50)) \\ Colin Barker, Sep 30 2018

From Colin Barker, Sep 30 2018: (Start)
G.f.: 3*x*(1 + x - 2*x^3 + 4*x^4 + 30*x^5 + x^6 - 5*x^7 + 24*x^8) / ((1 - x)^5*(1 + x + x^2)^4).
a(n) = a(n-1) + 4*a(n-3) - 4*a(n-4) - 6*a(n-6) + 6*a(n-7) + 4*a(n-9) - 4*a(n-10) - a(n-12) + a(n-13) for n>13.
(End)

A333446 Table T(n,k) read by upward antidiagonals. T(n,k) = Sum_{i=1..n} Product_{j=1..k} (i-1)*k+j.

Original entry on oeis.org

1, 3, 2, 6, 14, 6, 10, 44, 126, 24, 15, 100, 630, 1704, 120, 21, 190, 1950, 13584, 30360, 720, 28, 322, 4680, 57264, 390720, 666000, 5040, 36, 504, 9576, 173544, 2251200, 14032080, 17302320, 40320, 45, 744, 17556, 428568, 8626800, 110941200, 603353520, 518958720, 362880

Offset: 1

Chai Wah Wu, Mar 23 2020

T(n,k) is the maximum value of Sum_{i=1..n} Product_{j=1..k} r[(i-1)*k+j] among all permutations r of {1..kn}. For the minimum value see A331889.

			From _Seiichi Manyama_, Jul 23 2020: (Start)
T(3,2) = Sum_{i=1..3} Product_{j=1..2} (i-1)*2+j = 1*2 + 3*4 + 5*6 = 44.
Square array begins:
   1,   2,    6,     24,      120,        720, ...
   3,  14,  126,   1704,    30360,     666000, ...
   6,  44,  630,  13584,   390720,   14032080, ...
  10, 100, 1950,  57264,  2251200,  110941200, ...
  15, 190, 4680, 173544,  8626800,  538459200, ...
  21, 322, 9576, 428568, 25727520, 1940869440, ... (End)

Seiichi Manyama, Antidiagonals n = 1..140, flattened
Chai Wah Wu, On rearrangement inequalities for multiple sequences, arXiv:2002.10514 [math.CO], 2020.

Column k=1-3 give A000217, A268684, A268685(n-1).
Main diagonal gives A336513.
Cf. A323663, A331889,

def T(n,k): # T(n,k) for A333446
    c, l = 0, list(range(1,k*n+1,k))
    lt = list(l)
    for i in range(n):
        for j in range(1,k):
            lt[i] *= l[i]+j
        c += lt[i]
    return c

T(n,k) = Sum_{i=1..n} Gamma(ik+1)/Gamma((i-1)k+1).

cp's OEIS Frontend

A319014 a(n) = 123 + 456 + 789 + 101112 + 131415 + 161718 + ... + (up to n).

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A319867 a(n) = 321 + 654 + 987 + 121110 + ... + (up to the n-th term).

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A333446 Table T(n,k) read by upward antidiagonals. T(n,k) = Sum_{i=1..n} Product_{j=1..k} (i-1)*k+j.

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cp's OEIS Frontend

A319014 a(n) = 1*2*3 + 4*5*6 + 7*8*9 + 10*11*12 + 13*14*15 + 16*17*18 + ... + (up to n).

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A319867 a(n) = 3*2*1 + 6*5*4 + 9*8*7 + 12*11*10 + ... + (up to the n-th term).

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Maple

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A333446 Table T(n,k) read by upward antidiagonals. T(n,k) = Sum_{i=1..n} Product_{j=1..k} (i-1)*k+j.

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Comments

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Python

Formula

A319014 a(n) = 123 + 456 + 789 + 101112 + 131415 + 161718 + ... + (up to n).

A319867 a(n) = 321 + 654 + 987 + 121110 + ... + (up to the n-th term).