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Revision History for A339019

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A339019

Square table read by upwards antidiagonals: T(m,n) = A103438(2*m-1,n)/A103438(1,n) for m>=1, n>=1.
(history; published version)

#38 by N. J. A. Sloane at Sat Feb 06 22:33:30 EST 2021

STATUS

editing

approved

#37 by N. J. A. Sloane at Sat Feb 06 22:33:27 EST 2021

NAME

Square Table table read by upwards antidiagonals of : T(m,n) = A103438(2*m-1,n)/A103438(1,n) for m>=1, n>=1.

STATUS

proposed

editing

#36 by Alois P. Heinz at Mon Jan 04 17:08:27 EST 2021

STATUS

editing

proposed

#35 by Alois P. Heinz at Mon Jan 04 17:06:53 EST 2021

NAME

Square Table by upwards antidiagonals of T(m,n) = A103438(2*m-1,n)/A103438(1,n) for m>=1, n>=1.

KEYWORD

nonn,tabl,changed

STATUS

proposed

editing

#34 by Franz Vrabec at Mon Jan 04 15:52:04 EST 2021

STATUS

editing

proposed

#33 by Franz Vrabec at Mon Jan 04 15:51:29 EST 2021

EXAMPLE

T(3,4) = A103438(2*3-1,4)/A103438(1,4) = 1300 / 10 = 130.

CROSSREFS

Cf. A103438.

#32 by Franz Vrabec at Sun Jan 03 11:34:00 EST 2021

EXAMPLE

By formula: a(2,3) = 4*15*1*1*B(4) = -2 and a(3,3) = (-8)*15*4*(2/4)*B(4) = 8 yields T(3,n) = (-N+4*N^2)/3. Since N = 4*5/2 = 10, so T(3,4) = (4*10^2-10)/3 = 130.

#31 by Franz Vrabec at Mon Dec 28 14:42:21 EST 2020

EXAMPLE

T(3,4) = A103438(2*3-1,4)/A103438(1,4) = 1300 / 10 = 130.

T(3,n) = (4*N^2-N)/3 for N = Sum_{j=1..n} j, because a(2,3) = 4*15*1*1*B(4) = -2 and a(3,3) = (-8)*15*4*(2/4)*B(4) = 8 yields T(3,n) = (-N+4*N^2)/3. N = 4*5/2 = 10, so T(3,4) = (4*10^2-10)/3 = 130.

For n = 4 N = 10, so T(3,4) = (4*10^2-10)/3 = 130 = 1300 / 10 = A103438(2*3-1,4)/A103438(1,4)

1 | 1 1 1 1 1 1 1

2 | 1 3 6 10 15 21 28

3 | 1 11 46 130 295 581 1036

4 | 1 43 386 1870 6455 17941 42868

5 | 1 171 3366 28234 149031 586341 1880956

6 | 1 683 29866 437350 3546775 19809461 85475908

7 | 1 2731 267086 6871138 85960967 683338501 3972825676

#30 by Franz Vrabec at Sun Dec 27 15:38:00 EST 2020

DATA

1, 1, 1, 1, 3, 1, 1, 11, 6, 1, 1, 43, 46, 10, 1, 1, 171, 386, 130, 15, 1, 1, 683, 3366, 1870, 295, 21, 1, 1, 2731, 29866, 28234, 6455, 581, 28, 1, 1, 10923, 267086, 437350, 149031, 17941, 1036, 36, 1

EXAMPLE

For n = 4 N = 10, so T(3,4) = (4*10^2-10)/3 = 130 = 1300 / 10 = A103438(2*3-1,4)/A103438(1,4)

m\n| 1 2 3 4 5 6 7

---+----------------------------------------------

---+-----------------------------------------------------

1 | 1 1 1 1 1 1 1

2 | 1 3 6 10 15 21 28

3 | 1 11 46 130 295 581 1036

4 | 1 43 386 1870 6455 17941 42868

5 | 1 171 3366 28234 149031 586341 1880956

6 | 1 683 29866 437350 3546775 19809461 85475908

7 | 1 2731 267086 6871138 85960967 683338501 3972825676

#29 by Franz Vrabec at Sat Dec 26 14:12:54 EST 2020

FORMULA

Let a(i,km) = ((-2)^i)*sumSum_{j=0..i, } C(2*m,i-j)*C(i+j,j)*((i-j)/(i+j))*B(2*m-i+j), B(s) = A027641(s)/A027642(s) the Bernoulli numbers and N = n*(n+1)/2, then T(m,n) = (sum1/(2*m))*Sum_{i=12..m, } a(i,m)*N^(i-1))/(2*m)}.

EXAMPLE

T(3,n) = (4*N^2-N)/3 for N = Sum_{j=1..n} j, because a(2,3) = 4*15*1*1*B(4) = -2 and a(3,3) = (-8)*15*4*(2/4)*B(4) = 8.

1 | 1 1 1 1 1 1

2 | 1 3 6 10 15 21

3 | 1 11 46 130 295 581

4 | 1 43 386 1870 6455 17941

5 | 1 171 3366 28234 149031 586341

6 | 1 683 29866 437350 3546775 19809461