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User:Tobias O'Leary

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Switching careers from Software Develoent to Secondary math teacher. My former professor Dr. Gottlieb of Rhodes College gave me a link to this site when I found 1*n + 2*(n-1) + ... + n*1 = ( (n+1)^3 - n )/6.

0, 1, 4, 10, 20, 35, 56, 84, 120

a(n) = [(n+1)^3 - (n+1)]/6. Subtract the sum of positive squares from the sum of positive intergers multiples by n and you get a(n-1).

Home Address: 3113 Patterson Ave Apt 1 Richmond, VA 23221 USA