T(n)^2=T(T(n))+T(T(n)-1)
From specialfunctionswiki
Theorem
The following formula holds for $n=2,3,4,\ldots$: $$T(n)^2=T(T(n))+T(T(n)-1),$$ where $T(n)$ denotes the $n$th triangular number.
Proof
References
- {{ #if: |{{{2}}}|V.E. Hoggatt, Jr}}{{#if: Marjorie Bicknell|{{#if: |, {{ #if: |{{{2}}}|Marjorie Bicknell}}{{#if: |, [[Mathematician:{{{author3}}}|{{ #if: |{{{2}}}|{{{author3}}}}}]]{{#if: |, [[Mathematician:{{{author4}}}|{{ #if: |{{{2}}}|{{{author4}}}}}]]{{#if: |, [[Mathematician:{{{author5}}}|{{ #if: |{{{2}}}|{{{author5}}}}}]] and [[Mathematician:{{{author6}}}|{{ #if: |{{{2}}}|{{{author6}}}}}]]| and [[Mathematician:{{{author5}}}|{{ #if: |{{{2}}}|{{{author5}}}}}]]}}| and [[Mathematician:{{{author4}}}|{{ #if: |{{{2}}}|{{{author4}}}}}]]}}| and [[Mathematician:{{{author3}}}|{{ #if: |{{{2}}}|{{{author3}}}}}]]}}| and {{ #if: |{{{2}}}|Marjorie Bicknell}}}}|}}: [[Paper:V.E. Hoggatt, Jr/Triangular numbers{{#if: |/Volume {{{volume}}}|}}{{#if: |/{{{edpage}}}}}|Triangular numbers{{#if: |: Volume {{{volume}}}|}}{{#if: |: {{{eddisplay}}} (1974)| ({{#if: |{{{ed}}} ed., }}1974)}}]]{{#if: |, {{{publisher}}}|}}{{#if: |, ISBN {{{isbn}}}|}}{{#if: N^2=T(n)+T(n-1) | ... (previous)|}}{{#if: T(n+1)^2-T(n)^2=(n+1)^3 | ... (next)|}}{{#if: |: Entry: {{#if: |[[{{{entryref}}}|{{{entry}}}]]|{{{entry}}}}}|}} $(1.4)$
