1,2

Colm Mulcahy, Dec 2007 Card Colm

There are exactly 657 ways of splitting the numbers 1..16 into two groups so that the numbers in each group have the same sum s1=68.

There are exactly 57 ways of splitting the numbers 1..16 into two groups so that the numbers in each group have the same sum of squares s2=748.

And there is the only one way of splitting the numbers 1..16 into two groups so that the numbers in each group have the same sum of cubes s3=9248: {{1, 4, 6, 7, 10, 11, 13, 16} (=A133483) and {2, 3, 5, 8, 9, 12, 14, 15} (=A133484)}.

Amazingly, this last way also appears in previous two cases! This is given in A133483 and A133484.

(Also, there is the only one way of splitting the numbers 1..16 into two groups so that the numbers in each group have the same sum of fourth powers s4=121924: {{1,2,3,4,8,9,10,11,12,16},{5,6,7,13,14,15}}. But this splitting does not give equal sum of powers 1..3.)

Intersection of first two cases gives 12 ways of splitting the numbers 1..16 into two groups so that the numbers in each group have the same sum s1=68 and the same sum of squares s2=748;

Here we list only groups containing 1 (the corresponding 2nd groups are their complements):

{1,2,3,4,8,9,11,14,16},

{1,2,3,5,6,10,11,14,16},

{1,2,3,5,7,8,12,14,16},

{1,2,4,5,6,8,11,15,16},

{1,2,7,8,11,12,13,14},

{1,3,6,8,10,12,13,15},

{1,3,6,9,10,11,12,16},

{1,4,5,8,10,11,14,15},

{1,4,6,7,9,12,14,15},

{1,4,6,7,10,11,13,16},

{1,5,6,7,8,11,14,16}.

This table read by rows gives the present sequence.

Cf. A133483, A133484, A135418.

Sequence in context: A047546 A139759 A030292 this_sequence A051914 A104519 A117220

Adjacent sequences: A135416 A135417 A135418 this_sequence A135420 A135421 A135422

fini,full,nonn,tabf

Zak Seidov (zakseidov(AT)yahoo.com), Dec 01 2007

Error in example lines corrected by Colm Mulcahy (colm(AT)spelman.edu), Dec 30 2007