Bobby Milazzo - cp's OEIS frontend

User: Bobby Milazzo

Bobby Milazzo has authored 2 sequences.

A197083 Number of solutions to a+b+c = d+e+f with 0 < a <= n, 0 <= b,c,d,e,f <= n.

0, 10, 96, 445, 1431, 3681, 8141, 16142, 29466, 50412, 81862, 127347, 191113, 278187, 394443, 546668, 742628, 991134, 1302108, 1686649, 2157099, 2727109, 3411705, 4227354, 5192030, 6325280, 7648290, 9183951, 10956925, 12993711, 15322711, 17974296, 20980872

Offset: 0

Bobby Milazzo, Mar 11 2013

When n < 10, a(n) is the number of six-digit numbers (with digits <= n) that have the property that the sum of the rightmost 3 digits equals the sum of the leftmost 3 digits. Some references call these balanced numbers. [Edited by M. F. Hasler, Mar 11 2013]

			When n=1, a(n)=10 because there are 10 solutions when viewed as balanced numbers: 111111, 110110, 110101, 110011, 101110, 101101, 101011, 100100, 100010, 100001.

Index entries for linear recurrences with constant coefficients, signature (6,-15,20,-15,6,-1).

RecurrenceTable[{a[0] == 0, a[1] == 10, a[2] == 96, a[3] == 445, a[4] == 1431, a[5] == 3681, a[n] == 6 a[n - 1] - 15 a[n - 2] + 20 a[n - 3] - 15 a[n - 4] + 6 a[n - 5] - a[n - 6]}, a, {n, 0, 35}]

def A197083(n): return n*(n*(n*(n*(66*n+275)+440)+325)+94)//120 # Chai Wah Wu, May 08 2024

G.f.: (x^4 + 19*x^3 + 36*x^2 + 10*x)/(x-1)^6.
a(n) = 6*a(n-1) - 15*a(n-2) + 20*a(n-3) - 15*a(n-4) + 6*a(n-5) - a(n-6) for n > 5; a(0)=0, a(1)=10, a(2)=96, a(3)=445, a(4)=1431, a(5)=3681.
a(n) = (66*n^5 + 275*n^4 + 440*n^3 + 325*n^2 + 94*n)/120 = n*(n+1)*(66*n^3 + 209*n^2 + 231*n + 94)/120.

A179441 Number of solutions to a+b+c < d+e with each of a,b,c,d,e in {1..n+1}.

Original entry on oeis.org

1, 21, 121, 432, 1182, 2723, 5558, 10368, 18039, 29689, 46695, 70720, 103740, 148071, 206396, 281792, 377757, 498237, 647653, 830928, 1053514, 1321419, 1641234, 2020160, 2466035, 2987361, 3593331, 4293856, 5099592, 6021967, 7073208, 8266368, 9615353, 11134949, 12840849

Offset: 1

Bobby Milazzo, Jul 14 2010

			a(1) = 1 since from {1,2} there is only one solution: {1,1,1} for a,b,c and {2,2} for d,e.
a(2) = 21 since from {1,2,3} there are 21 ways to select a,b,c,d,e such that a+b+c < d+e.

Mathematics and Computer Education 1988 - 89 #261 Unsolved.

Andrew Howroyd, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (6,-15,20,-15,6,-1).

Cf. A197083.

k=10;
Table[p=Expand[Sum[x^k,{k,1,n}]^2 Sum[1/x^k,{k,1,n}]^3];
Twowins=Drop[CoefficientList[p,x],1]//Total,{n,2,k}]

a(n)=(27*n^5 + 80*n^4 + 65*n^3 - 20*n^2 - 32*n)/120 \\ Andrew Howroyd, Apr 15 2021

a(n) = (1/120)*(27*n^5 + 80*n^4 + 65*n^3 - 20*n^2 - 32*n).
a(n) = 6*a(n-1) - 15*a(n-2) + 20*a(n-3) - 15*a(n-4) + 6*a(n-5) - a(n-6) for n > 6.
G.f.: x*(1 + 15*x + 10*x^2 + x^3)/(1 - x)^6.

Name edited and terms a(24) and beyond from Andrew Howroyd, Apr 15 2021

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User: Bobby Milazzo

A197083 Number of solutions to a+b+c = d+e+f with 0 < a <= n, 0 <= b,c,d,e,f <= n.

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A179441 Number of solutions to a+b+c < d+e with each of a,b,c,d,e in {1..n+1}.

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