A347839 An array of the positive integers congruent to 2 modulo 3 (A016789), read by antidiagonals upwards, giving the present triangle.
2, 5, 8, 11, 20, 32, 14, 44, 80, 128, 17, 56, 176, 320, 512, 23, 68, 224, 704, 1280, 2048, 26, 92, 272, 896, 2816, 5120, 8192, 29, 104, 368, 1088, 3584, 11264, 20480, 32768, 35, 116, 416, 1472, 4352, 14336, 45056, 81920, 131072, 38, 140, 464, 1664, 5888, 17408, 57344, 180224, 327680, 524288
Offset: 1
Examples
The array a(k, n) begins: k \ n 0 1 2 3 4 5 6 7 8 9 10 ... --------------------------------------------------------------------------- 1: 2 8 32 128 512 2048 8192 32768 131072 524288 2097152 ... 2: 5 20 80 320 1280 5120 20480 81920 327680 1310720 5242880 ... 3: 11 44 176 704 2816 11264 45056 180224 720896 2883584 11534336 ... 4: 14 56 224 896 3584 14336 57344 229376 917504 3670016 14680064 ... 5: 17 68 272 1088 4352 17408 69632 278528 1114112 4456448 17825792 ... 6: 23 92 368 1472 5888 23552 94208 376832 1507328 6029312 24117248 ... 7: 26 104 416 1664 6656 26624 106496 425984 1703936 6815744 27262976 ... 8: 29 116 464 1856 7424 29696 118784 475136 1900544 7602176 30408704 ... 9: 35 140 560 2240 8960 35840 143360 573440 2293760 9175040 36700160 ... 10: 38 152 608 2432 9728 38912 155648 622592 2490368 9961472 39845888 ... ... ---------------------------------------------------------------------------- The triangle t(n,k) begins: k \ n 0 1 2 3 4 5 6 7 8 9 ... --------------------------------------------------------------- 1: 2 2: 5 8 3: 11 20 32 4: 14 44 80 128 5: 17 56 176 320 512 6: 23 68 224 704 1280 2048 7: 26 92 272 896 2816 5120 8192 8: 29 104 368 1088 3584 11264 20480 32768 9: 35 116 416 1472 4352 14336 45056 81920 131072 10: 38 140 464 1664 5888 17408 57344 180224 327680 524288 ... -----------------------------------------------------------------
Crossrefs
Programs
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Maple
A := (n, k) -> 4^n*(3*(k + iquo(k, 3)) - 1): for k from 1 to 10 do seq(A(n, k), n = 0..10) od; # Alternatively: gf := n -> (4^n*((z*(z*(7*z + 3) + 3) - 1)))/((z - 1)^2*(1 + z + z^2)): ser := n -> series(gf(n), z, 12): col := (n, len) -> seq(coeff(ser(n), z, k), k = 1..len): seq(print(col(n, 10)), n = 0..10); # Peter Luschny, Oct 26 2021
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Mathematica
A[n_, k_] := 4^n (3(k + Quotient[k, 3]) - 1); Table[A[n-k, k], {n, 1, 10}, {k, n, 1, -1}] // Flatten (* Jean-François Alcover, Nov 07 2021, from Maple code *)
Formula
Array a:
a(k, n) = (3*A(k, n) + 1)/2, with the array A from A347834, for k >= 1, and n >= 0.
a(k, n) = 4^n*A347838(k) = 4^n*(2 + 3*k + 3*floor((k + 1)/3)).
Recurrence for rows k: a(k, n) = 4*a(k, n-1), for n >= 1, with a(k, 0) = A347838(k).
O.g.f.: expansion in z gives the o.g.f.s for rows k, also for k = 0: -A000302; expansion in x gives the o.g.f.s for columns n.
G(z, x) = (-1 + 3*z + 3*z^2 + 7*z^3)/((1 - z)*(1 - z^3)*(1 - 4*x)).
Triangle t:
t(k, n) = a(k-n, n), for k >= 1, and n = 0, 1, ..., k-1.
Comments